Advertisements
Advertisements
Question
If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.
Solution
a + b + c = 0 ...(i)
⇒ (a + b) + c = 0
Cubing both sides
⇒ (a + b)3 + c3 + 3(a + b) (c) (a+ b + c) = 0
⇒ a3 + b3 + 3ab (a + b) + c3 + 0 = 0
⇒ a3 + b3 + c3 + 3ab (a + b) = 0 ...(2)
Using (i), we get, a + b = -c From (2),
a3 + b3 + c3 + 3ab (-c) = 0
⇒ a3 + b3 + c3 = 3abc.
APPEARS IN
RELATED QUESTIONS
Expand.
`(x + 1/x)^3`
Expand.
`((5x)/y + y/(5x))^3`
If a + 2b = 5; then show that : a3 + 8b3 + 30ab = 125.
If a ≠ 0 and `a- 1/a` = 3 ; Find :
`a^3 - 1/a^3`
If X ≠ 0 and X + `1/"X"` = 2 ; then show that :
`x^2 + 1/x^2 = x^3 + 1/x^3 = x^4 + 1/x^4`
If `"r" - (1)/"r" = 4`; find : `"r"^3 - (1)/"r"^3`
If a + 2b + c = 0; then show that a3 + 8b3 + c3 = 6abc
If p - q = -1 and pq = -12, find p3 - q3
Expand: (3x + 4y)3.
Expand (3p + 4q)3
