Advertisements
Advertisements
Question
If a ≠ 0 and `a - 1/a` = 4 ; find : `( a^3 - 1/a^3 )`
Sum
Advertisements
Solution
`(a - 1/a)^3 = a^3 - 1/a^3 - 3( a - 1/a )`
⇒ `( a^3 - 1/a^3 ) = (a - 1/a)^3 + 3( a - 1/a )`
⇒ `( a^3 - 1/a^3 ) = (4)^3 + 3(4)` [ ∵ `a - 1/a = 4` ]
⇒ `( a^3 - 1/a^3 ) = 64 + 12 `
⇒ `( a^3 - 1/a^3 ) = 76`
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Use property to evaluate:
133 + (–8)3 + (–5)3
If a ≠ 0 and `a - 1/a` = 3 ; find `a^2 + 1/a^2`
If a ≠ 0 and `a - 1/a` = 4 ; find : `( a^4 + 1/a^4 )`
If `("a" + 1/"a")^2 = 3`; then show that `"a"^3 + (1)/"a"^3 = 0`
If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
Simplify:
`("a" + 1/"a")^3 - ("a" - 1/"a")^3`
Evaluate the following :
(3.29)3 + (6.71)3
Expand: (41)3
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2
Expand (3p + 4q)3
