Advertisements
Advertisements
प्रश्न
If `a + 1/a` = p and a ≠ 0; then show that:
`a^3 + 1/a^3 = p(p^2 - 3)`
बेरीज
Advertisements
उत्तर
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
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
Expand.
(k + 4)3
Expand.
(52)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:
- Sum of these numbers
- Difference of their cubes
- Sum of their cubes.
If a ≠ 0 and `a - 1/a` = 3 ; find `a^2 + 1/a^2`
If a ≠ 0 and `a- 1/a` = 3 ; Find :
`a^3 - 1/a^3`
If `9"a"^2 + (1)/(9"a"^2) = 23`; find the value of `27"a"^3 + (1)/(27"a"^3)`
If `"r" - (1)/"r" = 4`; find : `"r"^3 - (1)/"r"^3`
Simplify:
`("a" + 1/"a")^3 - ("a" - 1/"a")^3`
Expand: (3x + 4y)3.
Expand (52)3
