Question

please answer this......
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Answer 1

Answer:

Step-by-step explanation:x3+y3+z3-3xyz= (x + y + z) (x2 +y2 + z2 yz zx xy)

x3+y3+z3=(x +y + z) (x2+y2+ z2yz zx xy) + 3xyz ( Rearranging the terms)

Putting the values,

x3+y3+z3= (1) {(x2+y2+ z2- ( -1)} +3 (-1)

x3+y3+z3= (1) (x2+y2+ z2+1 ) -3 ------------------- Equation1

Now,

x2+y2+ z2

(x+y+z)2 =x2+y2+ z2+ 2 (xy+yz+zx)

(1)2 =x2+y2+ z2+2(-1)

1 =x2+y2+ z2-2

3=x2+y2+ z2----------------------------- Equation 2

Putting values of 2 on 1, we get

x3+y3+z3= (1) (3+1 ) -3

x3+y3+z3= (1) (4) -3

x3+y3+z3 = 1

Answer 2

Answer:

$\huge\pink{hey\:mate}$

Hey mate please refer to the attachment