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Expansion of (a ± b)³

Question 34 F...

Question

Question 34
Factorise:
(i) $1 + 64 x^{3}$

(ii) $a^{3} - 2 \sqrt{2} b^{3}$

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Solution

(i) $1 + 64 x^{3} = (1)^{3} + (4 x)^{3}$
$= (1 + 4 x) [(1)^{2} - (1) (4 x) + (4 x)^{2}]$
$[U s i n g t h e i d e n t i t y, a^{3} - b^{3} = (a + b) (a^{2} - a b + b^{2})]$
$= (1 + 4 x) (1 - 4 x + 16 x^{2})$

(ii) $a^{3} - 2 \sqrt{2} b^{3} = (a)^{3} - (\sqrt{2} b)^{3}$
$= (a - \sqrt{2} b) [a^{2} + a (\sqrt{2} b) + (\sqrt{2} b)^{2}]$
$[U s i n g t h e i d e n t i t y, a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})]$
$(a - \sqrt{2} b) (a^{2} + \sqrt{2} a b + 2 b^{2})$

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Question 34
Factorise:
(i) $1 + 64 x^{3}$

(ii) $a^{3} - 2 \sqrt{2} b^{3}$

Question 36
Factorise:
(i) $a^{3} - 8 b^{3} - 64 c^{3} - 24 a b c$

(ii) $2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$

a^{3} - 2 \sqrt{2} b^{3}

Factorise:
$a^{3} + 3 a^{2} b + 3 a b^{2} + 2 b^{3}$

Q. Find which is the greater

3 a b^{2}

a^{3} + 2 b^{3}

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Expansion of (a ± b)³

Standard IX Mathematics

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