Advertisements
Advertisements
प्रश्न
Expand.
`(x + 1/x)^3`
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`
संबंधित प्रश्न
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
Use property to evaluate:
133 + (–8)3 + (–5)3
The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes.
If X ≠ 0 and X + `1/"X"` = 2 ; then show that :
`x^2 + 1/x^2 = x^3 + 1/x^3 = x^4 + 1/x^4`
Find the cube of: 2a - 5b
Find the cube of: `4"p" - (1)/"p"`
If `5x + (1)/(5x) = 7`; find the value of `125x^3 + (1)/(125x^3)`.
If p - q = -1 and pq = -12, find p3 - q3
Simplify:
`("a" + 1/"a")^3 - ("a" - 1/"a")^3`
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2
