Advertisements
Advertisements
प्रश्न
If `a + 1/a` = p and a ≠ 0; then show that:
`a^3 + 1/a^3 = p(p^2 - 3)`
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)`
APPEARS IN
संबंधित प्रश्न
Expand.
(7 + m)3
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
If `a^2 + 1/a^2` = 18; a ≠ 0 find :
(i) `a - 1/a`
(ii) `a^3 - 1/a^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:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
Find the cube of: 4x + 7y
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"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)`
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2
If `x^2 + 1/x^2` = 23, then find the value of `x + 1/x` and `x^3 + 1/x^3`
(p + q)(p2 – pq + q2) is equal to _____________