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प्रश्न
Factorise:
`16a^3 - 128/b^3`
योग
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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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
संबंधित प्रश्न
Simplify:
\[\frac{8 x^3 - 27 y^3}{4 x^2 - 9 y^2}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Factorise:
y3 − 27
Factorise:
8p3 −\[\frac{27}{p^3}\]
Simplify:
(a + b)3 − a3 − b3
Simplify:
p3 − (p + 1)3
Factorise: x3 - 8y3
Factorise: 54p3 - 250q3.
Factorise: `a^3 - 1/(a^3)`
Simplify: (2x + 3y)3 - (2x - 3y)3
