Q. यदि `x+y+z=6` , `xyz= -10` तथा `x^2+y^2+z^2 = 30` है , तो `x^3 +y^3+z^3` का मान क्या होगा ?
Solution by Studypanal.in
solve- we know that
Solved by Studypanal
`(x^3+y^3+z^3)-3xyz = (x+y+z) ( x^2+y^2+z^2-xy-yz-zx)`
`(x^3+y^3+z^3) = (x+y+z) ( x^2+y^2+z^2-xy-yz-zx)+3xyz`
first of all we have to find `xy-yz-zx`
`x+y+z=6`
On Squaring bothsides we have
`(x+y+z)^2 = ( x^2+y^2+z^2 + 2xy +2yz+2zx )`
`(6)^2 = 30 + 2 ( xy + yz +zx )`
`36 = 30 + 2 ( xy + yz + zx )`
`36-.30 = 2 ( xy + yz + zx )`
`6 = 2 ( xy + yz + zx )`
` ( xy + yz + zx )= 3`
Now,
`(x^3+y^3+z^3) = (x+y+z) ( x^2+y^2+z^2-xy-yz-zx)+3xyz`
`(x^3+y^3+z^3) = 6\times (30-3) + 3\times -10`
`(x^3+y^3+z^3)= 6 \times27 -30 `
`(x^3+y^3+z^3) = 162-30`
`(x^3+y^3+z^3)= 132`
thanks for visit our websites