Factorise- 8a3-27-b3-8

We know a^3 b^3 =(a b) (a^2 +ab +b^2) 8a^3/27 b^3/8 =( 2a/3)^3 ( b/2)^3 = ( 2a/3 b/2) (( 2a/3)^2 + 2a/3×b/2 +(b/2)^2) = ( 2a/3 b/2) (4a^2/9 +2ab/6 +b^2 ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation Using Algebraic Identities

Factorise: 8a...

Question

Factorise:

$\frac{8 a^{3}}{27} - \frac{b^{3}}{8}$

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Solution

$W e k n o w a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \frac{8 a^{3}}{27} - \frac{b^{3}}{8} = (\frac{2 a}{3})^{3} - (\frac{b}{2})^{3} = (\frac{2 a}{3} - \frac{b}{2}) ((\frac{2 a}{3})^{2} + \frac{2 a}{3} \times \frac{b}{2} + (\frac{b}{2})^{2}) = (\frac{2 a}{3} - \frac{b}{2}) (\frac{4 a^{2}}{9} + \frac{2 a b}{6} + \frac{b^{2}}{4})$

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Similar questions

Q. Factorize :

\frac{8 a^{3}}{27} - \frac{b^{3}}{8}

Factorise each of the following:

(i) (ii)

(iii) (iv)

(v)

Factorise:

$8 a^{3} - b^{3} - 4 a x + 2 b x$

Factorise : 8a³ + b³ + 12a²b + 6ab²

Q. Factorise

8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}

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Factorise: 8a327−b38

We know a3−b3=(a−b)(a2+ab+b2)8a327−b38=(2a3)3−(b2)3=(2a3−b2)((2a3)2+2a3×b2+(b2)2)=(2a3−b2)(4a29+2ab6+b24)

Factorise:

$\frac{8 a^{3}}{27} - \frac{b^{3}}{8}$