Advertisements
Advertisements
Question
If `a + 1/a` = p and a ≠ 0; then show that:
`a^3 + 1/a^3 = p(p^2 - 3)`
Sum
Advertisements
Solution
Given that `a + 1/a` = p ...(1)
`(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 = (p)^3 - 3(p)` ...[From(1)]
⇒ `a^3 + 1/a^3 = p(p^2 - 3)`
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Expand.
(52)3
Use property to evaluate : 73 + 33 + (-10)3
Find the cube of: 2a - 5b
Find the cube of: `(2"m")/(3"n") + (3"n")/(2"m")`
If `5x + (1)/(5x) = 7`; find the value of `125x^3 + (1)/(125x^3)`.
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"a"^3 - (1)/"a"^3`
Simplify:
(a + b)3 + (a - b)3
Evaluate the following :
(5.45)3 + (3.55)3
Evaluate the following :
(8.12)3 - (3.12)3
Expand: `[x + 1/y]^3`
