Sum and Difference of Two Cubes

We now look at two special results obtained from multiplying a binomial and a trinomial:

Sum of two cubes:

\begin{align*} (x + y)({x}^{2} - xy + {y}^{2}) & = x({x}^{2} - xy + {y}^{2}) + y({x}^{2} - xy + {y}^{2}) \\ & = [x({x}^{2}) + x(-xy) + x({y}^{2})] + [y({x}^{2}) + y(-xy) + y({y}^{2})]\\ & = {x}^{3} - {x}^{2}y + x{y}^{2} + {x}^{2}y -x{y}^{2} + {y}^{3}\\ & = {x}^{3} + {y}^{3} \end{align*}

Difference of two cubes:

\begin{align*} (x - y)({x}^{2} + xy + {y}^{2}) & = x({x}^{2} + xy + {y}^{2}) - y({x}^{2} + xy + {y}^{2})\\ & = [x({x}^{2}) + x(xy) + x({y}^{2})] - [y({x}^{2}) + y(xy) + y({y}^{2})]\\ & = {x}^{3} + {x}^{2}y + x{y}^{2} - {x}^{2}y - x{y}^{2} - {y}^{3}\\ & = {x}^{3} - {y}^{3} \end{align*}

So we have seen that:

\begin{align*} {x}^{3} + {y}^{3} & = (x + y)({x}^{2} - xy + {y}^{2}) \\ {x}^{3} - {y}^{3} & = (x - y)({x}^{2} + xy + {y}^{2}) \end{align*}

We use these two basic identities to factorise more complex examples.

Example

Question

Factorise: \({a}^{3}-1\).

Take the cube root of terms that are perfect cubes

We are working with the difference of two cubes. We know that \({x}^{3} - {y}^{3} = (x - y)({x}^{2} + xy + {y}^{2})\), so we need to identify \(x\) and \(y\).

We start by noting that \(\sqrt[3]{{a}^{3}}=a\) and \(\sqrt[3]{1}=1\). These give the terms in the first bracket. This also tells us that \(x = a\) and \(y = 1\).

Find the three terms in the second bracket

We can replace \(x\) and \(y\) in the factorised form of the expression for the difference of two cubes with \(a\) and \(\text{1}\). Doing so we get the second bracket:

\[({a}^{3} - 1) = (a - 1)({a}^{2} + a + 1)\]

Expand the brackets to check that the expression has been correctly factorised

\begin{align*} (a - 1)({a}^{2} + a + 1) & = a({a}^{2} + a + 1) - 1({a}^{2} + a + 1) \\ & = {a}^{3} + {a}^{2} + a - {a}^{2} - a - 1 \\ & = {a}^{3} - 1 \end{align*}

Example

Question

Factorise: \({x}^{3}+8\).

Take the cube root of terms that are perfect cubes

We are working with the sum of two cubes. We know that \({x}^{3} + {y}^{3} = (x + y)({x}^{2} - xy + {y}^{2})\), so we need to identify \(x\) and \(y\).

We start by noting that \(\sqrt[3]{{x}^{3}}=x\) and \(\sqrt[3]{8}=2\). These give the terms in the first bracket. This also tells us that \(x = x\) and \(y = 2\).

Find the three terms in the second bracket

We can replace \(x\) and \(y\) in the factorised form of the expression for the sum of two cubes with \(x\) and \(\text{2}\). Doing so we get the second bracket:

\[({x}^{3} + 8) = (x + 2)({x}^{2} - 2x + 4)\]

Expand the brackets to check that the expression has been correctly factorised

\begin{align*} (x + 2)({x}^{2} - 2x + 4) & = x({x}^{2} - 2x + 4) + 2({x}^{2} - 2x + 4) \\ & = {x}^{3} - 2{x}^{2} + 4x + 2{x}^{2} - 4x + 8 \\ & = {x}^{3} + 8 \end{align*}

Example

Question

Factorise: \(16{y}^{3}-432\).

Take out the common factor 16

\[16{y}^{3} - 432 = 16({y}^{3} - 27)\]

Take the cube root of terms that are perfect cubes

We are working with the difference of two cubes. We know that \({x}^{3} - {y}^{3} = (x - y)({x}^{2} + xy + {y}^{2})\), so we need to identify \(x\) and \(y\).

We start by noting that \(\sqrt[3]{{y}^{3}}=y\) and \(\sqrt[3]{27}=3\). These give the terms in the first bracket. This also tells us that \(x = y\) and \(y = 3\).

Find the three terms in the second bracket

We can replace \(x\) and \(y\) in the factorised form of the expression for the difference of two cubes with \(y\) and \(\text{3}\). Doing so we get the second bracket:

\[16({y}^{3} - 27) = 16(y - 3)({y}^{2} + 3y + 9)\]

Expand the brackets to check that the expression has been correctly factorised

\begin{align*} 16(y - 3)({y}^{2} + 3y + 9) & = 16[(y({y}^{2} + 3y + 9) - 3({y}^{2} + 3y + 9)] \\ & = 16[{y}^{3} + 3{y}^{2} + 9y - 3{y}^{2} - 9y - 27] \\ & = 16{y}^{3} - 432 \end{align*}

Example

Question

Factorise: \(8{t}^{3}+125{p}^{3}\).

Take the cube root of terms that are perfect cubes

We are working with the sum of two cubes. We know that \({x}^{3} + {y}^{3} = (x + y)({x}^{2} - xy + {y}^{2})\), so we need to identify \(x\) and \(y\).

We start by noting that \(\sqrt[3]{{8t}^{3}}=2t\) and \(\sqrt[3]{125p^{3}}=5p\). These give the terms in the first bracket. This also tells us that \(x = 2t\) and \(y = 5p\).

Find the three terms in the second bracket

We can replace \(x\) and \(y\) in the factorised form of the expression for the difference of two cubes with \(2t\) and \(5p\). Doing so we get the second bracket:

\begin{align*} (8{t}^{3} + 125{p}^{3}) & = (2t + 5p)[{(2t)}^{2} - (2t)(5p) + {(5p)}^{2}] \\ & = (2t + 5p)(4{t}^{2} - 10tp + 25{p}^{2}) \end{align*}

Expand the brackets to check that the expression has been correctly factorised

\begin{align*} (2t + 5p)(4{t}^{2} - 10tp + 25{p}^{2}) & = 2t(4{t}^{2} - 10tp + 25{p}^{2}) + 5p(4{t}^{2} - 10tp + 25{p}^{2}) \\ & = 8{t}^{3} - 20p{t}^{2} + 50{p}^{2}t + 20p{t}^{2} - 50{p}^{2}t + 125{p}^{3} \\ & =8{t}^{3} + 125{p}^{3} \end{align*}

Sum and Difference of Two Cubes