Exploring Algebraic Identities

In every part, identify \(a\) and \(b\) first, then write down the three terms \(a^2\), \(2ab\) and \(b^2\).

(i) Here \(a=7x\) and \(b=4y\).

(ii) Here \(a=\frac{7}{5}x\) and \(b=\frac{3}{2}y\). The middle term is \(2\times\frac{7}{5}\times\frac{3}{2}=\frac{21}{5}\).

(iii) Here \(a=2.5p\) and \(b=1.5q\), so \(2ab = 2(2.5)(1.5)pq = 7.5pq\).

(iv) Here \(a=\frac{3}{4}s\) and \(b=8t\), so \(2ab = 2\times\frac{3}{4}\times 8 = 12\).

(v) Here \(a=x\) and \(b=\frac{1}{2y}\), so \(2ab = 2\times x\times\frac{1}{2y}=\frac{x}{y}\).

(vi) Here \(a=\frac{1}{x}\) and \(b=\frac{1}{y}\).

Split each number into a convenient round part plus a small part, then apply \((a+b)^2=a^2+2ab+b^2\).

(i) Write \(64 = 60+4\).

(ii) Write \(105 = 100+5\).

(iii) Write \(205 = 200+5\).

Each expression is to be matched with \(a^2+2ab+b^2=(a+b)^2\). Take the square roots of the first and last terms to guess \(a\) and \(b\), then check the middle term equals \(2ab\).

(i) \(\sqrt{9x^2}=3x\), \(\sqrt{16y^2}=4y\) and \(2(3x)(4y)=24xy\). It matches.

(ii) \(\sqrt{4s^2}=2s\), \(\sqrt{25t^2}=5t\) and \(2(2s)(5t)=20st\).

(iii) \(\sqrt{49x^2}=7x\), \(\sqrt{4y^2}=2y\) and \(2(7x)(2y)=28xy\).

(iv) \(\sqrt{64p^2}=8p\), \(\sqrt{\tfrac{4}{9}q^2}=\tfrac{2}{3}q\) and \(2(8p)\!\left(\tfrac{2}{3}q\right)=\tfrac{32}{3}pq\).

*(v) The first term \(3a^2\) is not a perfect square, so follow the hint and pull out a common factor first — take out 3:

Inside the bracket \(\sqrt{a^2}=a\), \(\sqrt{\tfrac{4}{9}b^2}=\tfrac{2}{3}b\) and \(2(a)\!\left(\tfrac{2}{3}b\right)=\tfrac{4}{3}ab\).

*(vi) Again the leading coefficient is not a perfect square, so take out \(\frac{9}{5}\):

Each number is just short of a round number, so write it as a difference.

(i) \(79 = 80-1\).

(ii) \(193 = 200-7\).

(iii) \(299 = 300-1\).

Choose the split that leaves the smallest, roundest pieces. A number close below a round number suits \((a-b)^2\); a three-digit number with three significant digits suits \((a+b+c)^2\).

(i) \(117 = 100+10+7\), so use \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\).

(ii) \(78 = 80-2\), so use \((a-b)^2\).

(iii) \(198 = 200-2\), so use \((a-b)^2\).

(iv) \(214 = 200+10+4\), so use \((a+b+c)^2\).

(v) \(1104 = 1100+4\), so use \((a+b)^2\).

(vi) \(1120 = 1100+20\), so use \((a+b)^2\).

(i) Two perfect squares with a negative middle term point to \((a-b)^2\). Here \(a=4y\), \(b=3\) and \(2ab=24y\).

(ii) \(\sqrt{\tfrac{9}{4}s^2}=\tfrac{3}{2}s\), \(\sqrt{4t^2}=2t\) and \(2\!\left(\tfrac{3}{2}s\right)(2t)=6st\).

(iii) Six terms with three perfect squares means the three-variable identity \((a+b+c)^2\). The squares are \(\tfrac{m^2}{9}\), \(\tfrac{k^2}{4}\) and \(9n^2\), so try \(a=\tfrac{m}{3}\), \(b=\tfrac{k}{2}\), \(c=3n\).

All three cross terms match, so:

(iv) Take \(a=\tfrac{p}{4}\) and \(b=\tfrac{4}{p}\). Then \(2ab = 2\times\tfrac{p}{4}\times\tfrac{4}{p}=2\), which is exactly the middle term with its minus sign.

(v) Three squares again. Try \(a=3a\), \(b=-2b\), \(c=c\); the signs of the cross terms tell you which letter is negative.

(i) Take \(a=p\), \(b=3q\), \(c=7r\).

(ii) Take \(a=3x\), \(b=-2y\), \(c=4z\) and keep the signs inside the products.

An identity must hold for every value of the letters, so a single counter-example is enough to settle it. Try the simplest possible choice, \(a=b=c=1\).

Left side. Each bracket becomes:

Right side.

Since \(3 \neq 6\), the statement is not an identity.

What the left side actually equals. Expanding each bracket in full:

Adding the three lines, the \(2ab\) terms give \(-2ab\), the \(2bc\) terms give \(-2bc\) and the \(2ca\) terms give \(-2ca\):

This equals \(2a^2+2b^2+2c^2\) only for special values of \(a\), \(b\) and \(c\) — not for all of them. So it is an equation, not an identity.

(i) Compare with \(s^2+(a+b)s+ab\): we need \(a+b=-11\) and \(ab=24\). The pair \(-3\) and \(-8\) works.

(ii) One factor is \((x+1)\). Since the product starts with \(3x^2\) and ends with \(-7\), the other factor must start with \(3x\) and end with \(-7\).

So the blank is \((3x-7)\).

(iii) Write the factors as \((2x-\alpha)(\beta x+2)\). Multiplying out, the \(x^2\) term is \(2\beta x^2 = 10x^2\), so \(\beta=5\); the constant is \(-2\alpha=-6\), so \(\alpha=3\). Check the middle term:

So the blanks are \(3\) and \(5x\).

(iv) Here \(6x^2+7x+2\) needs two numbers multiplying to \(6\times 2=12\) and adding to \(7\) — namely \(3\) and \(4\). Split the middle term:

Squares of numbers near a round number use \((a\pm b)^2\). A product of two numbers equally spaced about a round number uses \((a+b)(a-b)=a^2-b^2\). Any other product of two numbers uses \((x+a)(x+b)=x^2+(a+b)x+ab\).

(i) \(41 = 40+1\), so use \((a+b)^2\).

(ii) \(27 = 30-3\), so use \((a-b)^2\).

(iii) \(23\) and \(17\) are both 3 away from 20, so use \(a^2-b^2\).

(iv) \(135 = 130+5\), so use \((a+b)^2\).

(v) \(97 = 100-3\), so use \((a-b)^2\).

(vi) \(18\) and \(29\) are not equally spaced about any round number, so use \((x+a)(x+b)\) with \(x=20\), \(a=-2\), \(b=9\).

(vii) Take \(x=40\), \(a=-6\), \(b=3\).

(viii) \(205 = 200+5\), so use \((a+b)^2\).

(i) Three squares and three cross terms — use \((a+b+c)^2\). The squares are \(9a^2\), \(b^2\) and \(4c^2\), and the signs force \(b\) to be the negative one. Try \(3a\), \(-b\), \(2c\):

(ii) Rearranged, this is \(16s^2-40st+25t^2\), with \(a=4s\), \(b=5t\) and \(2ab=40st\).

(iii) Find two numbers with product \(-42\) and sum \(-1\): they are \(-7\) and \(6\).

(iv) \(\sqrt{49g^2}=7g\), \(\sqrt{h^2}=h\) and \(2(7g)(h)=14gh\).

(v) The squares are \(64u^2\), \(121v^2\) and \(4w^2\). Two cross terms are negative and one positive, so both \(v\) and \(w\) carry a minus. Try \(8u\), \(-11v\), \(-2w\):

The method is always the same: factor the numerator, factor the denominator, then cancel whatever is common.

(i) Factor both parts.

Look carefully: the numerator's factors are \(p-3q\) and \(p+2q\), the denominator's are \(p+5q\) and \(p-2q\). No factor is common, so nothing cancels — the expression is already in its simplest form.

(ii) The numerator is \((n-m)^3\) and the denominator is \(5(m-n)^2 = 5(n-m)^2\).

(iii) The denominator is a perfect square, \((w-v+x)^2\). For the numerator use \(a^3+b^3+c^3-3abc\) with \(a=w\), \(b=-v\), \(c=x\); then \(-3abc = -3(w)(-v)(x) = +3wvx\), which is exactly the last term.

Cancelling one factor of \((w-v+x)\):

(iv) The numerator is \((2y-5z)^2\) and the denominator is a difference of squares.

Since \((2y-5z)^2 = (5z-2y)^2\), one factor of \((5z-2y)\) cancels:

(v) Factor all four quadratics.

Every factor upstairs appears downstairs, so all four cancel:

(vi) The numerator is a difference of squares twice over; the denominator is \((p-2)^2\).

(i) Use \((a+b)^2\) with \(a=-3x\), \(b=4\).

(ii) Use \((a+b)(a-b)=a^2-b^2\).

(iii) Same identity, with \(a=p^2\) and \(b=\tfrac{1}{2}\).

(iv) Again \(a^2-b^2\), with \(a=2n\), \(b=7\).

(v) This is \((a-b)(a^2+ab+b^2)=a^3-b^3\) with \(a=s\), \(b=2t\).

(vi) Use \((a-b)^2\) with \(a=\tfrac{1}{2r}\), \(b=4r\); the middle term is \(2\times\tfrac{1}{2r}\times 4r = 4\).

(vii) Use \((a+b+c)^2\) with \(a=-3m\), \(b=4k\), \(c=-l\).

(viii) Use \((a-b)^3 = a^3-3a^2b+3ab^2-b^3\) with \(a=x\), \(b=\tfrac{y}{3}\).

(ix) Same identity with \(a=\tfrac{7}{2}k\), \(b=\tfrac{2}{3}m\).

Parts (i)–(iii) are pairs equally spaced about a round number, so use \(a^2-b^2\). Parts (iv)–(viii) are cubes near a round number, so use \((a\pm b)^3\).

(i) \(17\) and \(21\) are both 2 away from 19.

(ii) Both are 4 away from 100.

(iii) Both are 4 away from 20.

(iv) \(147 = 150-3\), so use \((a-b)^3 = a^3-3a^2b+3ab^2-b^3\).

(v) \(199 = 200-1\).

(vi) \(127 = 130-3\).

(vii) A negative cube stays negative, so compute \(107^3\) with \(107 = 100+7\) and put the sign back.

(viii) Similarly \(299 = 300-1\).

(i) Two squares with middle term 1. Take \(a=2y\), \(b=\tfrac{1}{4y}\); then \(2ab = 2(2y)\!\left(\tfrac{1}{4y}\right)=1\).

(ii) A difference of squares.

(iii) A difference of cubes, \(a^3-b^3=(a-b)(a^2+ab+b^2)\), with \(a=3b\) and \(b'=\tfrac{1}{4b}\).

(iv) Need two numbers with sum \(\tfrac{5}{6}\) and product \(\tfrac{1}{6}\) — they are \(\tfrac{1}{2}\) and \(\tfrac{1}{3}\).

(v) Four terms with alternating signs suggest \((a-b)^3\). Take \(a=3u\), \(b=\tfrac{1}{5}\).

(vi) A sum of cubes, \(a^3+b^3=(a+b)(a^2-ab+b^2)\), with \(a=4y\), \(b=\tfrac{z}{5}\).

(vii) This is \(a^3+b^3+c^3-3abc\) with \(a=p\), \(b=3q\), \(c=r\), since \(3abc = 3(p)(3q)(r)=9pqr\).

(viii) A perfect square with \(a=3m\), \(b=2\).

(ix) None of \(9x^3\), \(\tfrac{8}{3}y^3\), \(\tfrac{z^3}{3}\) is a clean cube, so take \(\tfrac{1}{3}\) out as a common factor first.

Inside the bracket use \(a^3+b^3+c^3-3abc\) with \(a=3x\), \(b=-2y\), \(c=z\); then \(-3abc = -3(3x)(-2y)(z) = 18xyz\).

(x) The three squares \(4x^2\), \(9y^2\) and \(36z^2\) point to \(a=2x\), \(b=3y\), \(c=6z\). Their cross terms are:

(xi) Again \((a-b)^3\), this time with \(a=3u\) and \(b=\tfrac{1}{6}\).

(i) The numerator is \((2x+1)^2\); the denominator is a difference of squares.

(ii) Pull out the numbers first, then use the difference of cubes and the difference of squares.

(iii) The numerator is a sum of cubes with \(a=s\), \(b=5t\); the denominator factors as a quadratic in \(s\).

Area of a rectangle is length × breadth, so factoring the area into two factors gives one possible pair of dimensions.

(i) This is a perfect square with \(a'=5a\), \(b'=3b\) and \(2a'b' = 30ab\).

So length \(= 5a-3b\) units and breadth \(= 5a-3b\) units — the rectangle is in fact a square.

(ii) A difference of squares.

So length \(= 6s+7t\) units and breadth \(= 6s-7t\) units.

A cuboid needs three factors, so factor the volume into three parts.

(i) Take out 6, then use the difference of squares.

So the three dimensions can be \(6\), \((a-2b)\) and \((a+2b)\) units.

(ii) Take out the common factor \(3p\), then factor the quadratic.

So the three dimensions can be \(3p\), \((s-1)\) and \((s-4)\) units.

The path forms a border all the way round, so the outer boundary is also a square. Its side is the playground's side plus the path width on both ends.

The area of the path is the outer square minus the playground:

Expand using \((a+b)^2\):

It can also be written in factored form as \(4s(s+40)\) square metres.

Let the number be \(x\). Its reciprocal is \(\frac{1}{x}\), so:

Multiply throughout by \(3x\) to clear the fractions:

Split the middle term. Two numbers with product \(3\times 3 = 9\) and sum \(-10\) are \(-1\) and \(-9\):

So \(x = \frac{1}{3}\) or \(x = 3\).

Check. Both work, and they are reciprocals of each other:

The number is 3 (or equivalently \(\frac{1}{3}\)).

Length \(=\) Area \(\div\) Width, so factor the area and divide out the known width.

Split the middle term of \(2x^2+7x+3\): two numbers with product \(2\times 3=6\) and sum \(7\) are \(1\) and \(6\).

Dividing by the width \((2x+1)\):

The length of the pool is \((x+3)\) hastas.

If \((x-k)\) is a factor of a polynomial, then substituting \(x=k\) makes the polynomial zero.

Using the factor \(x-2\), put \(x=2\):

Using the factor \(x-\frac{1}{2}\), put \(x=\frac{1}{2}\):

Multiply this second equation by 4 to clear the fraction:

Now both equations have the same right-hand side, so their left-hand sides are equal:

which is what had to be shown.

Going further. Substituting \(p=r\) back into \(4p+r=-10\) gives \(5p=-10\), so \(p=r=-2\) and the polynomial is \(-2x^2+5x-2 = -(2x-1)(x-2)\).

Start from the identity established in this chapter:

The bracket contains \(a^2+b^2+c^2\), which is not given directly — but it follows from \((a+b+c)^2\):

Now substitute both known quantities into the identity:

Hence proved.

Factor the expression completely.

The bracket is a difference of squares:

Rewriting the factors in increasing order:

This is the product of three consecutive integers. Now argue about divisibility:

Divisible by 2. Among any two consecutive integers one is even, so among three consecutive integers at least one is even. Hence the product is divisible by 2.

Divisible by 3. Every third integer is a multiple of 3, so among any three consecutive integers exactly one is a multiple of 3. Hence the product is divisible by 3.

Since 2 and 3 are coprime, a number divisible by both is divisible by \(2\times 3 = 6\). Therefore \(n^3-n\) is always divisible by 6.

Both parts are disguised versions of \(a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\). The trick is to spot the third cube hiding in the constant term.

(i) Note that \(64 = 4^3\), and \(-12xy = -3(x)(y)(4)\). So take \(a=x\), \(b=y\), \(c=4\).

Since \(x+y=-4\), the first bracket is:

A product with a zero factor is zero, so the value is 0.

(ii) Here \(-8y^3 = (-2y)^3\), \(-216 = (-6)^3\), and \(-3(x)(-2y)(-6) = -36xy\). So take \(a=x\), \(b=-2y\), \(c=-6\).

Therefore the expression factors as:

Since \(x = 2y+6\), the first bracket is:

So the value is 0.