Question

Factorise:
(1) y³ − 27
(2) x³ − 64y³
(3) 27m³ − 216n³
(4) 125y³ − 1
(5) 8p³ − $\frac{27}{p^3}$
(6) 343a³ − 512b³
(7) 64x³ − 729y³
(8) 16a³ − $\frac{128}{b^3}$

Answer 1

It is known that,
$a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)$
$$\begin{gathered} \left(1\right)y^3-27 \\ ={\left(y\right)}^3-{\left(3\right)}^3 \\ =\left(y-3\right)\left\{{\left(y\right)}^2+{\left(3\right)}^2+\left(y\right)\times \left(3\right)\right\} \\ =\left(y-3\right)\left(y^2+9+3y\right) \end{gathered}$$
$$\begin{gathered} \left(2\right)x^3-64y^3 \\ ={\left(x\right)}^3-{\left(4y\right)}^3 \\ =\left(x-4y\right)\left\{{\left(x\right)}^2+{\left(4y\right)}^2+\left(x\right)\times \left(4y\right)\right\} \\ =\left(x-4y\right)\left(x^2+16y^2+4xy\right) \end{gathered}$$
$$\begin{gathered} \left(3\right)27m^3-216n^3 \\ =27\left[m^3-8n^3\right] \\ =27\left[{\left(m\right)}^3-{\left(2n\right)}^3\right] \\ =27\left[\left(m-2n\right)\left\{{\left(m\right)}^2+{\left(2n\right)}^2+\left(m\right)\times \left(2n\right)\right\}\right] \\ =27\left(m-2n\right)\left(m^2+4n^2+2mn\right) \end{gathered}$$
$$\begin{gathered} \left(4\right)125y^3-1 \\ ={\left(5y\right)}^3-{\left(1\right)}^3 \\ =\left(5y-1\right)\left\{{\left(5y\right)}^2+{\left(1\right)}^2+\left(5y\right)\times \left(1\right)\right\} \\ =\left(5y-1\right)\left(25y^2+1+5y\right) \end{gathered}$$
$$\begin{gathered} \left(5\right)8p^3-\frac{27}{p^3} \\ ={\left(2p\right)}^3-{\left(\frac{3}{p}\right)}^3 \\ =\left(2p-\frac{3}{p}\right)\left\{{\left(2p\right)}^2+{\left(\frac{3}{p}\right)}^2+\left(2p\right)\times \left(\frac{3}{p}\right)\right\} \\ =\left(2p-\frac{3}{p}\right)\left(4p^2+\frac{9}{p^2}+6\right) \end{gathered}$$
$$\begin{gathered} \left(6\right)343a^3-512b^3 \\ ={\left(7a\right)}^3-{\left(8b\right)}^3 \\ =\left(7a-8b\right)\left\{{\left(7a\right)}^2+{\left(8b\right)}^2+\left(7a\right)\times \left(8b\right)\right\} \\ =\left(7a-8b\right)\left(49a^2+64b^2+56ab\right) \end{gathered}$$
$$\begin{gathered} \left(7\right)64x^3-729y^3 \\ ={\left(4x\right)}^3-{\left(9y\right)}^3 \\ =\left(4x-9y\right)\left\{{\left(4x\right)}^2+{\left(9y\right)}^2+\left(4x\right)\times \left(9y\right)\right\} \\ =\left(4x-9y\right)\left(16x^2+81y^2+36xy\right) \end{gathered}$$
$$\begin{gathered} \left(8\right)16a^3-\frac{128}{b^3} \\ =16\left[a^3-\frac{8}{b^3}\right] \\ =16\left[{\left(a\right)}^3-{\left(\frac{2}{b}\right)}^3\right] \\ =16\left[\left(a-\frac{2}{b}\right)\left\{{\left(a\right)}^2+{\left(\frac{2}{b}\right)}^2+\left(a\right)\times \left(\frac{2}{b}\right)\right\}\right] \\ =16\left(a-\frac{2}{b}\right)\left(a^2+\frac{4}{b^2}+\frac{2a}{b}\right) \end{gathered}$$