Advertisements
Advertisements
Question
Expand: `[x + 1/y]^3`
Sum
Advertisements
Solution
(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
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Expand.
(7 + m)3
Find the cube of: `( 3a - 1/a ) (a ≠ 0 )`
If a + 2b + c = 0; then show that: a3 + 8b3 + c3 = 6abc.
If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.
If a ≠ 0 and `a- 1/a` = 3 ; Find :
`a^3 - 1/a^3`
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`
If `"a"^2 + (1)/"a"^2 = 14`; find the value of `"a"^3 + (1)/"a"^3`
If `9"a"^2 + (1)/(9"a"^2) = 23`; find the value of `27"a"^3 + (1)/(27"a"^3)`
If x3 + y3 = 9 and x + y = 3, find xy.
Expand: (x + 3)3.
