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
Expand.
(k + 4)3
Use property to evaluate : 73 + 33 + (-10)3
Find the cube of: 4x + 7y
If `5x + (1)/(5x) = 7`; find the value of `125x^3 + (1)/(125x^3)`.
If `"m"^2 + (1)/"m"^2 = 51`; find the value of `"m"^3 - (1)/"m"^3`
If `"r" - (1)/"r" = 4`; find : `"r"^3 - (1)/"r"^3`
If p - q = -1 and pq = -12, find p3 - q3
If m - n = -2 and m3 - n3 = -26, find mn.
If x + 2y = 5, then show that x3 + 8y3 + 30xy = 125.
Expand: (41)3
