Advertisements
Advertisements
प्रश्न
Expand: `[x + 1/y]^3`
बेरीज
Advertisements
उत्तर
(a + b)3 = a3 + b3 + 3ab(a + b)
`[x + 1/y]^3 = x^3 + 1/y^3 + 3 xx x xx 1/y(x + 1/y)`
= `x^3 + 1/y^3 + (3x)/y(x + 1/y)`
= `x^3 + 1/y^3 + (3x^2)/y + (3x)/(y^2)`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
If a2 + `1/a^2 = 47` and a ≠ 0 find :
- `a + 1/a`
- `a^3 + 1/a^3`
Expand : (3x - 5y - 2z) (3x - 5y + 2z)
If a ≠ 0 and `a- 1/a` = 3 ; Find :
`a^3 - 1/a^3`
If a ≠ 0 and `a - 1/a` = 4 ; find : `( a^3 - 1/a^3 )`
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"a"^3 - (1)/"a"^3`
If `"a" + (1)/"a" = "p"`; then show that `"a"^3 + (1)/"a"^3 = "p"("p"^2 - 3)`
Simplify:
`("a" + 1/"a")^3 - ("a" - 1/"a")^3`
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2
Expand (52)3
Find the volume of the cube whose side is (x + 1) cm
