Factorise using suitable identities:
$27 a^{3} + 8 b^{3} - 27 c^{3} + 18 a b c$

Open in App

Solution

${27 a}^{3} + 8 b^{3} - 27 c^{3} + 18 a b c$ is of the form $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

${(3 a)}^{3} + {(2 b)}^{3} + {(- 3 c)}^{3} - 3 \times 3 a \times 2 b \times - 3 c$

$= (3 a + 2 b - 3 c) ({(3 a)}^{2} + {(2 b)}^{2} + {(- 3 c)}^{2} - 3 a \times 2 b - 2 b \times - 3 c - (- 3 c) \times 3 a)$

$= (3 a + 2 b - 3 c) ({9 a}^{2} + {4 b}^{2} + {9 c}^{2} - 6 a b + 6 b c + 9 a c)$

Suggest Corrections

Similar questions

27 a^{3} + b^{3} + 8 c^{3} - 18 a b c

If a+2b+3c=0,prove that a³+8b³+27c³=18abc

27 a^{3} + 8 b^{3} - 18 a^{2} b - 12 a b^{2}

{(a + 2 b - 3 c)}^{3} - a^{3} - 8 b^{3} + 27 c^{3}

2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c

Join BYJU'S Learning Program