Advertisements
Advertisements
प्रश्न
Factorise:
`16a^3 - 128/b^3`
योग
Advertisements
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)`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
संबंधित प्रश्न
Simplify:
\[\frac{m^2 - n^2}{\left( m + n \right)^2} \times \frac{m^2 + mn + n^2}{m^3 - n^3}\]
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
Factorise:
27m3 − 216n3
Factorise:
125y3 − 1
Factorise:
64x3 − 729y3
Simplify:
(a + b)3 − a3 − b3
Factorise: 27p3 - 125q3.
Factorise: 54p3 - 250q3.
Factorise: `a^3 - 1/(a^3)`
Factorise the following:
a6 – 64
