Question

Factorise.
(1) x³ + 64y³
(2) 125p³ + q³
(3) 125k³ + 27m³
(4) 2l³ + 432m³
(5) 24a³ + 81b³
(6) y³ + $\frac{1}{8y^3}$
(7) a³ + $\frac{8}{a^3}$
(8) 1 + $\frac{q^3}{125}$

Answer 1

It is known that,
$a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)$
$$\begin{gathered} \left(1\right)x^3+64y^3 \\ ={\left(x\right)}^3+{\left(4y\right)}^3 \\ =\left(x+4y\right)\left\{{\left(x\right)}^2+{\left(4y\right)}^2-\left(x\right)\times \left(4y\right)\right\} \\ =\left(x+4y\right)\left(x^2+16y^2-4xy\right) \end{gathered}$$
$$\begin{gathered} \left(2\right)125p^3+q^3 \\ ={\left(5p\right)}^3+{\left(q\right)}^3 \\ =\left(5p+q\right)\left\{{\left(5p\right)}^2+{\left(q\right)}^2-\left(5p\right)\times \left(q\right)\right\} \\ =\left(5p+q\right)\left(25p^2+q^2-5pq\right) \end{gathered}$$
$$\begin{gathered} \left(3\right)125k^3+27m^3 \\ ={\left(5k\right)}^3+{\left(3m\right)}^3 \\ =\left(5k+3m\right)\left\{{\left(5k\right)}^2+{\left(3m\right)}^2-\left(5k\right)\times \left(3m\right)\right\} \\ =\left(5k+3m\right)\left(25k^2+9m^2-15km\right) \end{gathered}$$
$$\begin{gathered} \left(4\right)2l^3+432m^3 \\ =2\left[l^3+216m^3\right] \\ =2\left[{\left(l\right)}^3+{\left(6m\right)}^3\right] \\ =2\left[\left(l+6m\right)\left\{{\left(l\right)}^2+{\left(6m\right)}^2-\left(l\right)\times \left(6m\right)\right\}\right] \\ =2\left(l+6m\right)\left(l^2+36m^2-6lm\right) \end{gathered}$$
$$\begin{gathered} \left(5\right)24a^3+81b^3 \\ =3\left[8a^3+27b^3\right] \\ =3\left[{\left(2a\right)}^3+{\left(3b\right)}^3\right] \\ =3\left[\left(2a+3b\right)\left\{{\left(2a\right)}^2+{\left(3b\right)}^2-\left(2a\right)\times \left(3b\right)\right\}\right] \\ =3\left(2a+3b\right)\left(4a^2+9b^2-6ab\right) \end{gathered}$$
$$\begin{gathered} \left(6\right)y^3+\frac{1}{8y^3} \\ ={\left(y\right)}^3+{\left(\frac{1}{2y}\right)}^3 \\ =\left(y+\frac{1}{2y}\right)\left\{{\left(y\right)}^2+{\left(\frac{1}{2y}\right)}^2-\left(y\right)\times \left(\frac{1}{2y}\right)\right\} \\ =\left(y+\frac{1}{2y}\right)\left(y^2+\frac{1}{4y^2}-\frac{1}{2}\right) \end{gathered}$$
$$\begin{gathered} \left(7\right)a^3+\frac{8}{a^3} \\ ={\left(a\right)}^3+{\left(\frac{2}{a}\right)}^3 \\ =\left(a+\frac{2}{a}\right)\left\{{\left(a\right)}^2+{\left(\frac{2}{a}\right)}^2-\left(a\right)\times \left(\frac{2}{a}\right)\right\} \\ =\left(a+\frac{2}{a}\right)\left(a^2+\frac{4}{a^2}-2\right) \end{gathered}$$
$$\begin{gathered} \left(8\right)1+\frac{q^3}{125} \\ ={\left(1\right)}^3+{\left(\frac{q}{5}\right)}^3 \\ =\left(1+\frac{q}{5}\right)\left\{{\left(1\right)}^2+{\left(\frac{q}{5}\right)}^2-\left(1\right)\times \left(\frac{q}{5}\right)\right\} \\ =\left(1+\frac{q}{5}\right)\left(1+\frac{q^2}{25}-\frac{q}{5}\right) \end{gathered}$$