Advertisements
Advertisements
प्रश्न
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 − 3abc =
विकल्प
108
207
669
729
उत्तर
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
संबंधित प्रश्न
Factorise the following:
8a3 – b3 – 12a2b + 6ab2
What are the possible expressions for the dimensions of the cuboids whose volume is given below?
| Volume : 12ky2 + 8ky – 20k |
Evaluate the following using identities:
(1.5x2 − 0.3y2) (1.5x2 + 0.3y2)
If `x^2 + 1/x^2 = 66`, find the value of `x - 1/x`
Evaluate of the following:
(9.9)3
Find the following product:
\[\left( \frac{x}{2} + 2y \right) \left( \frac{x^2}{4} - xy + 4 y^2 \right)\]
If \[x + \frac{1}{x} = 2\], then \[x^3 + \frac{1}{x^3} =\]
If a − b = −8 and ab = −12, then a3 − b3 =
Use the direct method to evaluate the following products :
(3x – 2y) (2x + y)
Use the direct method to evaluate the following products :
(5a + 16) (3a – 7)
Expand the following:
(m + 8) (m - 7)
Find the squares of the following:
3p - 4q2
Find the squares of the following:
(2a + 3b - 4c)
Evaluate, using (a + b)(a - b)= a2 - b2.
4.9 x 5.1
If p2 + q2 + r2 = 82 and pq + qr + pr = 18; find p + q + r.
If `"r" - (1)/"r" = 4`; find : `"r"^4 + (1)/"r"^4`
Simplify:
(x + 2y + 3z)(x2 + 4y2 + 9z2 - 2xy - 6yz - 3zx)
Factorise the following:
16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz
Expand the following:
(3a – 2b)3
Expand the following:
`(1/x + y/3)^3`
