Advertisements
Advertisements
प्रश्न
Find the cube of: `( 3a - 1/a ) (a ≠ 0 )`
योग
Advertisements
उत्तर
(a − b)3 = a3 − 3a2b + 3ab2 − b3
`(3a - 1/a)^3`
= `(3a)^3 - 3 × (3a)^2 × 1/a + 3. 3a (1/a)^2 - (1/a)^3`
= `27a^3 - 3 . 9a^2. 1/a + 9a. 1/a^2 - 1/a^3`
= `27a^3 - 27a + 9/a - 1/a^3`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Simplify.
(3r − 2k)3 + (3r + 2k)3
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
Use property to evaluate : 133 + (-8)3 + (-5)3
If 4x2 + y2 = a and xy = b, find the value of 2x + y.
The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes.
Find the cube of: `3"a" + (1)/(3"a")`
Find the cube of: `"a" - (1)/"a" + "b"`
If `x^2 + (1)/x^2 = 18`; find : `x^3 - (1)/x^3`
If `"a" + (1)/"a" = "p"`; then show that `"a"^3 + (1)/"a"^3 = "p"("p"^2 - 3)`
Expand: (41)3
