Question

Use algebraic division to show the below important factorisation:

$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

Accepted Answer

We have to prove, $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

$\Rightarrow \frac{x^{3} - y^{3}}{(x - y)} = (x^{2} + x y + y^{2}) \Rightarrow (x^{3} - y^{3}) \div (x - y) = (x^{2} + x y + y^{2})$

Now using the algebraic division method,

$(x^{3} - y^{3}) \div (x - y)$

Divide the first term of the polynomial by $x$ :

$x^{3} \div (x) = x^{2}$ .

Multiply $(x - y)$ by $x^{2}$ : $x^{2} (x - y) = x^{3} - x^{2} y$

Subtract: $(x^{3} - y^{3}) - (x^{3} - x^{2} y) = x^{2} y - y^{3}$

$\begin{matrix} x^{2} \\ x - y x^{3} - y^{3} \\ \frac{x^{3} - x^{2} y}{x^{2} y - y^{3}} \end{matrix}$

Divide $x^{2} y$ by $x$ : $x^{2} y \div x = x y$ .

Multiply $(x - y)$ by $x y$ : $(x y) (x - y) = x^{2} y - x y^{2}$

Subtract: $(x^{2} y - y^{3}) - (x^{2} y - x y^{2}) = x y^{2} - y^{3}$

$\begin{matrix} x^{2} + x y \\ x - y x^{3} - y^{3} \\ \frac{x^{3} - x^{2} y}{x^{2} y - y^{3}} \\ \frac{x^{2} y - x y^{2}}{x y^{2} - y^{3}} \end{matrix}$

Divide $x y^{2}$ by $x$ : $x y^{2} \div x = y^{2}$ .

Multiply $(x - y)$ by $y^{2}$ : $(y^{2}) (x - y) = x y^{2} - y^{3}$

Subtract: $(x y^{2} - y^{3}) - (x y^{2} - y^{3}) = 0$

$\begin{matrix} x^{2} + x y + y^{2} \\ x - y x^{3} - y^{3} \\ \frac{x^{3} - x^{2} y}{x^{2} y - y^{3}} \\ \frac{x^{2} y - x y^{2}}{x y^{2} - y^{3}} \\ \frac{x y^{2} - y^{3}}{0} \end{matrix}$

We know that, dividend $=$ divisor $\times$ quotient $+$ remainder

Here, $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2}) + 0$

$\Rightarrow x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

Hence, proved.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 5: EXERCISE 1D

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 5: EXERCISE 1D

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