Question

Solve This : Mods And Stars...

Verify that:

$x {}^{3} +y {}^{3} +z {}^{3} –3xyz \\ = ( \frac{1}{2} ) (x+y+z)[(x–y) {}^{2} +(y–z) {}^{2} +(z–x) {}^{2} ]$

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Answer 1

Question :-

Verify that

x³+y³+z³-3xyz = ½ (x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Solution :-

Required to prove :-

x³+y³+z³-3xyz = ½ ( x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Proof :

Consider RHS;

½ ( x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Identities frequently used :

• (a+b)² = a²+b²+2ab
• (a-b)² = a²+b²-2ab

→

½ (x+y+z) [(x²+y²-2xy)+(y²+z²-2yz)+(x²+z²-2xz)]

½ (x+y+z) [(2x²+2y²+2z²-2xy-2yz-2xz)]

½ (x+y+z) 2 [x²+y²+z²-xy-yz-xz]

(x+y+z) (x²+y²+z²-xy-yz-xz)

x(x²+y²+z²-xy-yz-xz)+y(x²+y²+z²-xy-yz-xz)+z(x²+y²+z²-xy-yz-xz)

(x³+xy²+xz²-x²y-xyz-x²z)+(x²y+y³+yz²-xy²-y²x-xyz)+(x²z+y²z+z³-xyz-yz²-xz² )

→ x³+y³+z³-3xyz

Hence,

LHS = RHS

Bonus tip :-

If a + b + c = 0 then,

a³ + b³ + c³ = 3abc ✓

This can be used to solve many questions <<

Answer 2

Answer:

☆Answered by Rohith kumar.R.

●Given:-

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

●To prove:-

LHS=RHS

●Proof:-

Solving RHS

=1 (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]

2

By using the formula of

●(a-b)^2=a^2+b^2-2ab

= 1 (x+y+z)[(x^2+y^2-2xy)+(y^2+z^2-2yz)

2

+( z^2+x^2-2zx)

= 1 (x+y+z)[(2x^2+2y^2+2z^2

2

-2xy-2yz-2zx)

= 1 (x+y+z)2 ( x^2+y^2+z^2-xy-yz-zx)

2

= (x+y+z) ( x^2+y^2+z^2-xy-yz-zx)

We already know that

x^3+y^3+z^3-3xyz= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

☆Therefore,

LHS=RHS .

☆Hence proved.

☆Hope it helps u mate.

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☆Thank you.