Advertisements
Advertisements
Question
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 − 3abc =
Options
108
207
669
729
Advertisements
Solution
We have to find the value of `a^3 +b^3 +c^3 - 3abc`
Given `a+b+c = 9,ab +bc +ca = 23`
Using identity `(a+b+c)^2 = a^2 +b^2 +c^2 +2ab +2bc + 2ca` we get,
`(9)^2 = a^2 +b^2 +c^2 +2 (ab+bc +ca)`
` 9 xx 9 = a^2 +b^2 +c^2 +2 xx 23`
`81 = a^2 +b^2 +c^2 +46`
By transposing +46 to left hand side we get,
`81-46 = a^2 +b^2 +c^2`
`35 = a^2 +b^2 +c^2`
Using identity `a^3 +b^3 +c^3 -3abc = (a+b+c)[a^2 + b^2 +c^2 - (ab+bc+ca)]`
`9 xx [35 -23]`
` = 9 xx 12`
` = 108`
The value of `a^3 +b^3 +c^3 -3abc` is 108.
APPEARS IN
RELATED QUESTIONS
Expand the following, using suitable identity:
(2x – y + z)2
Expand the following, using suitable identity:
(3a – 7b – c)2
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Evaluate the following using identities:
117 x 83
If \[x + \frac{1}{x} = 5\], find the value of \[x^3 + \frac{1}{x^3}\]
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{x}{7} + \frac{y}{3} \right) \left( \frac{x^2}{49} + \frac{y^2}{9} - \frac{xy}{21} \right)\]
If x = −2 and y = 1, by using an identity find the value of the following
If \[a^2 + \frac{1}{a^2} = 102\] , find the value of \[a - \frac{1}{a}\].
Use the direct method to evaluate :
`("a"/2-"b"/3)("a"/2+"b"/3)`
Evaluate, using (a + b)(a - b)= a2 - b2.
999 x 1001
Evaluate, using (a + b)(a - b)= a2 - b2.
4.9 x 5.1
If x + y = 9, xy = 20
find: x - y
If x + y = 9, xy = 20
find: x2 - y2.
If m - n = 0.9 and mn = 0.36, find:
m2 - n2.
If x + y = 1 and xy = -12; find:
x2 - y2.
If `"r" - (1)/"r" = 4`; find: `"r"^2 + (1)/"r"^2`
Simplify:
(x + y - z)2 + (x - y + z)2
Simplify:
(2x + y)(4x2 - 2xy + y2)
Simplify:
(x + 2y + 3z)(x2 + 4y2 + 9z2 - 2xy - 6yz - 3zx)
Using suitable identity, evaluate the following:
9992
