If 8 a 3-27 b 3-0then 4 a2-9 b2-A- 0B- 6 abC- 18abD- -6 ab

The correct option is B 6ab Using identity (x+y)^3= x^3 + y^3 + 3(xy)(x + y) we get ⇒ (2a + 3b)^3 = (8a^3 + 27b^3) + 3(2a)(3b)(2a + 3b) ⇒ (2a + 3b)^3 = 0 + 18a ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x+y)^3

If 8 a 3+27 b...

Question

If $(8 a^{3} + 27 b^{3}) = 0$

then

$(4 a^{2} + 9 b^{2}) =$

$0$

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$6 a b$

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$18 a b$

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$- 6 a b$

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Solution

The correct option is B
$6 a b$

Using identity $(x + y)^{3} = x^{3} + y^{3} + 3 (x y) (x + y)$ we get

$\Rightarrow (2 a + 3 b)^{3} = (8 a^{3} + 27 b^{3}) + 3 (2 a) (3 b) (2 a + 3 b)$

$\Rightarrow (2 a + 3 b)^{3} = 0 + 18 a b (2 a + 3 b) Given: (8 a^{3} + 27 b^{3}) = 0$

$\Rightarrow (2 a + 3 b)^{3} - 18 a b (2 a + 3 b) = 0$

$\Rightarrow (2 a + 3 b) ((2 a + 3 b)^{2} - 18 a b) = 0 Taking (2 a + 3 b) common.$

$∴ (2 a + 3 b)^{2} - 18 a b = 0$

$\Rightarrow 4 a^{2} + 9 b^{2} + 12 a b - 18 a b = 0$

$\Rightarrow 4 a^{2} + 9 b^{2} - 6 a b = 0$

$\Rightarrow 4 a^{2} + 9 b^{2} = 6 a b$

Suggest Corrections

If (8a3+27b3)=0 then (4a2+9b2)=

If $(8 a^{3} + 27 b^{3}) = 0$

then

$(4 a^{2} + 9 b^{2}) =$