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प्रश्न
Factorise:
`16a^3 - 128/b^3`
बेरीज
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उत्तर
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
`16a^3 - 128/b^3`
= `16[a^3 - 8/b^3]`
= `16[(a)^3 - (2/b)^3]`
= `16[(a - 2/b) {(a)^2 + (a) xx (2/b) + (2/b)^2}]`
= `16(a - 2/b) (a^2 + (2a)/b + 4/b^2)`
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या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
संबंधित प्रश्न
Simplify:
\[\frac{8 x^3 - 27 y^3}{4 x^2 - 9 y^2}\]
Simplify:
\[\frac{m^2 - n^2}{\left( m + n \right)^2} \times \frac{m^2 + mn + n^2}{m^3 - n^3}\]
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Simplify:
`(1 - 2x + x^2)/(1 - x^3) xx (1 + x + x^2)/(1 + x)`
Factorise:
y3 − 27
Factorise:
64x3 − 729y3
Factorise: `a^3 - 1/(a^3)`
Simplify: (a - b)3 - (a3 - b3)
Factorise the following:
27x3 – 8y3
