Advertisements
Advertisements
प्रश्न
Expand.
`(x/3 - 3/x)^3`
योग
Advertisements
उत्तर
`(x/3 - 3/x)^3`
(a − b)3 = a3 − 3a2b + 3ab2 − b3
= `(x/3)^3 - 3 xx (x/3)^2 xx (3/x) + 3 xx (x/3) xx (3/x)^2 - (3/x)^3`
= `(x^3)/27 - 3 xx (x^2)/9 xx 3/x + 3 xx x/3 xx 9/(x^2) - 27/(x^3)`
= `(x^3)/27 - (3x)/9 xx 3/1 + 3/3 xx 9/x - 27/(x^3)`
= `(x^3)/27 - (3x)/3 + 9/x - 27/(x^3)`
= `(x^3)/27 - x + 9/x - 27/(x^3)`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
संबंधित प्रश्न
Expand.
(2m − 5)3
Expand.
(198)3
Expand.
`(2p - 1/(2p))^3`
Expand (4p - 5q)3
Find the cube of 99 using the expansion formula.
If `(y - 1/y)^3` = 27, then find the value of `y^3 - 1/y^3`
Expand (2x – 4y)3
Expand (ab – c)3
(a – b) = 3 and ab = 5 then a3 – b3 = __________
Factorise : 3x3 – 45x2y + 225xy2 – 375y3
