Advertisements
Advertisements
प्रश्न
Expand: `[x + 1/y]^3`
योग
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Expand.
(52)3
Expand.
`(x + 1/x)^3`
Find the cube of : 5a + 3b
Use property to evaluate : 73 + 33 + (-10)3
Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.
If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
If a + b = 5 and ab = 2, find a3 + b3.
Expand: (x + 3)3.
Expand: `((2m)/n + n/(2m))^3`.
