Advertisements
Advertisements
Question
Expand.
`(x + 1/x)^3`
Sum
Advertisements
Solution
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
Is there an error in this question or solution?
RELATED QUESTIONS
Expand.
(7x + 8y)3
Find the cube of: `( 3a - 1/a ) (a ≠ 0 )`
Use property to evaluate : 383 + (-26)3 + (-12)3
If a ≠ 0 and `a - 1/a` = 4 ; find : `( a^4 + 1/a^4 )`
Find the cube of: `3"a" + (1)/(3"a")`
Find the cube of: `"a" - (1)/"a" + "b"`
If `("a" + 1/"a")^2 = 3`; then show that `"a"^3 + (1)/"a"^3 = 0`
If x + 2y = 5, then show that x3 + 8y3 + 30xy = 125.
Expand: `((2m)/n + n/(2m))^3`.
If `x^2 + 1/x^2` = 23, then find the value of `x + 1/x` and `x^3 + 1/x^3`
