Question

(a) If x+y+z=0, show that xcube+ycube+zcube=3xyz.
(b) show that (a-b)cube+(b-c) cube+(c-a) cube=3(a-b)(b-c)(c-a)
answer fast as possible.​

Answer 1

Is it true that $\bold{x+y+z=0}$ then $\bold{x^3+y^3+z^3=3xyz}$?

To show this, we have to simplify the given equality.

→ Use transposition to simplify

$\bold{x^3+y^3+z^3-3xyz=0}$

→ Identity applies here

$\bold{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0}$

→ Find the solution

$\bold{x+y+z=0}$ or $\bold{x^2+y^2+z^2-xy-yz-zx=0}$

If we substitute the first solution, it satisfies the second. Hence shown.

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Now let's move on to the last question.

We can solve the question based on the first one.

→ Take substitution.

• $\bold{x=a-b}$
• $\bold{y=b-c}$
• $\bold{z=c-a}$

$\bold{x+y+z=0}$ so $\bold{x^3+y^3+z^3=3xyz}$

→ Take substitution again. Hence shown.

$\bold{\therefore (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)}$

I hope it helps. The main part is identity, so I recommend you to know the identity.

Answer 2

Question= If x+y+z=0, show that xcube+ycube+zcube=3xyz.

(b) show that (a-b)cube+(b-c) cube+(c-a) cube=3(a-b)(b-c)(c-a)

Answer⬇️

we know

a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

we have,

x=a(b-c) or x/a=(b-c)

y=b(c-a) or y/b=(c-a)

z=c(a-b) or z/c=(a-b)

now,

(x/a)^3+(y/b)^3+(z/c)^3–3(x/a)(y/b)(z/c)

=(x/a+y/b+z/c){(x/a)^2+(y/b)^2+(z/c)^2-(x/a)(y/b)-(y/b)(z/c)-(z/c)(x/a)}

={(b-c)+(c-a)+(a-b)}{ same}

={b-c+c-a+a-b}{ same }

=0{ same }=0

so

(x/a)^3+(y/b)^3+(z/c)^3=3(x/a)(y/b)(z/c)

=3xyz/abc proved