Advertisements
Advertisements
प्रश्न
Find the cube of: `"a" - (1)/"a" + "b"`
Advertisements
उत्तर
Using (a + b + c)3
= a3 + b3 + c3 + 3a2b + 3a2c + 3b2a + 3c2a + 6abc
`("a" - 1/"a" + "b")`
= `"a"^3 + (-1/"a")^3 + "b"^3 + 3"a"^2(-1/"a") + 3"a"^2 + 3(-1/"a")^2 "b" + 3(-1/"a")^2 "a" + 3"b"^2"a" + 3"b"^2(-1/"a") + 6"a"(-1/"a")"b"`
= `"a"^3 - (1)/"a"^3 + "b"^3 - 3"a" + 3"a"^2"b" + (3"b")/"a"^2 + (3)/"a" + 3"b"^2"a" - (3"b"^2)/"a" - 6"b"`.
APPEARS IN
संबंधित प्रश्न
Expand.
`(x + 1/x)^3`
Expand.
`((5x)/y + y/(5x))^3`
If `a + 1/a` = p and a ≠ 0; then show that:
`a^3 + 1/a^3 = p(p^2 - 3)`
The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes.
If `9"a"^2 + (1)/(9"a"^2) = 23`; find the value of `27"a"^3 + (1)/(27"a"^3)`
If `"a" + (1)/"a" = "p"`; then show that `"a"^3 + (1)/"a"^3 = "p"("p"^2 - 3)`
If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
Expand: (41)3
Expand (3p + 4q)3
Expand (52)3
