Question

if X + Y + Z is equal to 10 and X square + Y square + z square is equal to 40 find X Y + Y z + zx and x cube + y cube + Z cube - 3 x y z​

Answer 1

Answer:

XY + YZ + ZX = 30

$X^{3}+Y^{3}+Z^{3}-3XYZ = 100$

Step-by-step explanation:

Given that-

X + Y + Z = 10

$X^{2} +Y^{2} + Z^{2} = 40$

We know that-

$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

So, $(X+Y+Z)^{2} = X^{2}+Y^{2}+Z^{2}+2(XY+YZ+ZX)$

$(10)^{2} = 40 + 2(XY+YZ+ZX)$

100 = 40 +  2(XY+YZ+ZX)

100 - 40 = 2(XY +YZ +ZX)

60 = 2(XY +YZ +ZX)

XY + YZ + ZX = 30

We know that-

$(a+b+c)^{3} = (a^{3}+b^{3}+c^{3}) + 3[(a+b+c)(ab+bc+ca)-abc]$

So, $(X+Y+Z)^{3} = (X^{3}+Y^{3}+Z^{3}) + 3[(X+Y+Z)(XY+YZ+ZX)-3XYZ]$

$(10)^{3} = X^{3}+Y^{3}+Z^{3} + 3[(10)(30)-XYZ]$

$1000 = X^{3}+Y^{3}+Z^{3}+3(300-XYZ)$

$1000 = X^{3}+Y^{3}+Z^{3}+900 - 3XYZ$

$1000 - 900 = X^{3}+Y^{3}+Z^{3}-3XYZ$

So, $X^{3}+Y^{3}+Z^{3}-3XYZ = 100$