Question

Find the value of x3+y3+z3
-3xyz, if x+y+z =12 and x2+y2+z2 = 70

Answer 1

Answer:jcjlhcklwey

Step-by-step explanation:

Answer 2

$\huge\star\bold\red{ANSWER}$

$\star\bold\red{Given:-}$ x+y+z = 12, x²+y²+z²= 70

$\star\bold\red{To}$$\bold\red{Find:-}$ x³+y³+z³-3xyz=?

$\star\bold\red{Solution:-}$

As we know,

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)-----(1)

(x+y+z)²=x²+y²+z²+2xy+2yz+2zx

=>(12)²=70+2xy+2yz+2zx

=>144=70(2xy+2yz+2zx)

=>2(xy+yz+zx)=144-70

=>xy+yz+zx=$\frac{74}{2}$

=>xy+yz+zx=37

put above value in equ-----(1)

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

=>(12)[70-1(xy+yz+zx)]

=>(12)[70-1(37)]

=>12×(70-37)

=>12×33=>396$\bold\red{....ans...}$

HOPE THIS HELPS YOU

MARK IT AS BRAINILIST

$\huge\star\bold\red{THANK}$$\huge\star\bold\red{YOU}$