Advertisements
Advertisements
प्रश्न
Expand.
`(x + 1/x)^3`
बेरीज
Advertisements
उत्तर
Here, a = x, b = `1/x`
We know that,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
∴ `(x + 1/x)^3 = x^3 + 3(x)^2(1/x) + 3(x)(1/x)^2 + (1/x)^3`
= `x^3 + 3x^2 xx 1/x + 3x xx 1/x^2 + 1/x^3`
= `x^3 + 3x + 3/x + 1/x^3`
∴ `(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
संबंधित प्रश्न
Expand.
(k + 4)3
Expand.
(7 + m)3
If `( a + 1/a )^2 = 3 "and a ≠ 0; then show:" a^3 + 1/a^3 = 0`.
If a ≠ 0 and `a- 1/a` = 3 ; Find :
`a^3 - 1/a^3`
If a ≠ 0 and `a - 1/a` = 4; find: `(a^2 + 1/a^2)`
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"a"^3 - (1)/"a"^3`
If `x^2 + (1)/x^2 = 18`; find : `x^3 - (1)/x^3`
If x3 + y3 = 9 and x + y = 3, find xy.
Evaluate the following :
(3.29)3 + (6.71)3
Expand (3 + m)3
