Factorise. (i) $125 p^{3} + q^{3}$ (ii) $24 a^{3} + 81 b^{3}$

Open in App

Solution

(i) $125 p^{3} + q^{3}$
$= (5 p)^{3} + q^{3}$
Here, a = 5p and b = q
$∴ 125 p^{3} + q^{3} = (5 p + q) [(5 p)^{2} - (5 p) (q) + q^{2}]$
... $[∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})]$
$= (5 p + q) (25 p^{2} - 5 p q + q^{2})$

(ii) $24 a^{3} + 81 b^{3}$
[Taking out the common factor 3]
$= 3 [(2 a)^{3} + (3 b)^{3}]$
Here, A =2a and B = 3b
$∴ 24 a^{3} + 81 b^{3}$
$= 3 {\begin{matrix} (2 a + 3 b) [((2 a)^{2} - (2 a) (3 b) + (3 b)^{2}] \end{matrix}}$
... $[∵ A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})]$
$= 3 (2 a + 3 b) (4 a^{2} - 6 a b + 9 b^{2})$

Suggest Corrections

Similar questions

\frac{1}{8 y^{3}}

\frac{8}{a^{3}}

\frac{q^{3}}{125}

24 a^{3} + 81 b^{3}

24 a^{3} + 81 b^{3}

x^{3} + 64 y^{3}

24 a^{3} + 81 b^{3}

125 p^{3} + q^{3}

Join BYJU'S Learning Program