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प्रश्न
Factorise:
x3 − 64y3
योग
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उत्तर
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
x3 − 64y3
= (x) − (4y)3
= (x − 4y)[(x)2 + (4xy) + (4y)2]
= (x − 4y)(x2 + 4xy + 16y2)
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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
संबंधित प्रश्न
Simplify:
\[\frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}\]
Simplify:
\[\frac{3 x^2 - x - 2}{x^2 - 7x + 12} \div \frac{3 x^2 - 7x - 6}{x^2 - 4}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Factorise:
8p3 −\[\frac{27}{p^3}\]
Factorise:
343a3 − 512b3
Factorise:
64x3 − 729y3
Simplify:
(a + b)3 − a3 − b3
Simplify:
p3 − (p + 1)3
Simplify: (2x + 3y)3 - (2x - 3y)3
Factorise the following:
27x3 – 8y3
