Question

Using the identity prove: $x^3+y^3+z^3-3xyz=(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$.

Answer 1

Hint: To prove the equation we will use the identity of $(a+b{)}^3=a^3+b^3+3ab(a+b)$. We can also write the equation as $a^3+b^3=(a+b{)}^3-3ab(a+b)$. Put this identity equation in the LHS of the equation $x^3+y^3+z^3-3xyz$, simplify the equation, use multiplication of polynomials. Use the identity of $(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2xz$ to simplify it further. Arrange the terms to obtain the RHS of the equation $(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$.

Complete step-by-step answer:
Here we have to prove the given equation, which is $x^3+y^3+z^3-3xyz=(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$.
We will use the identity to prove the LHS of the equation equals the RHS of the equation.
As we know the equation $(a+b{)}^3=a^3+b^3+3ab(a+b)$, we can write the equation as $a^3+b^3=(a+b{)}^3-3ab(a+b)$ also. We will use this identity equation to prove.
Now taking Left hand side of the equation,
$LHS=x^3+y^3+z^3-3xyz$
Now using the identity $a^3+b^3=(a+b{)}^3-3ab(a+b)$, we can write as $x^3+y^3=(x+y{)}^3-3xy(x+y)$.
Putting this in the above equation, $LHS=(x+y{)}^3-3xy(x+y)+z^3-3xyz$,
Arranging the terms of equation, $LHS=(x+y{)}^3+z^3-3xy(x+y)-3xyz$
Now again using the same identity for $(x+y{)}^3+z^3$, we can write the equation as $(x+y{)}^3+z^3={\left\{(x+y)+z\right\}}^3-3(x+y)z\left\{(x+y)+z\right\}$.
Putting the equation of $(x+y{)}^3+z^3$ in our LHS equation, $LHS={\left\{(x+y)+z\right\}}^3-3(x+y)z\left\{(x+y)+z\right\}-3xy(x+y)-3xyz$
So, $$LHS=(x+y+z{)}^3-3(x+y)z(x+y+z)-3xy(x+y)-3xyz$$
Use multiplication of two polynomials to simplify the equation. For multiplication of two polynomials, we multiply each monomial of one polynomial with its sign by each monomial of another polynomial with its sign.
So, $$LHS=(x+y+z{)}^3-3(x+y)z(x+y+z)-3xyx-3xyy-3xyz$$
Taking $-3xy$ common in last three terms above equation,
$$LHS=(x+y+z{)}^3-3(x+y)z(x+y+z)-3xy(x+y+z)$$
Now taking $(x+y+z)$ common in all terms of the equation,
So, $$LHS=(x+y+z)\left\{{(x+y+z)}^2-3(x+y)z-3xy\right\}$$
Now use another identity of $(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2xz$. Put this in the above equation.
So, $$LHS=(x+y+z)\left\{(x^2+y^2+z^2+2xy+2yz+2xz)-3(x+y)z-3xy\right\}$$,
Multiply the polynomials, $$LHS=(x+y+z)\left\{x^2+y^2+z^2+2xy+2yz+2xz-3xz-3yz-3xy\right\}$$
Arranging the terms, $$LHS=(x+y+z)\left\{x^2+y^2+z^2+2xy-3xy+2yz-3yz+2xz-3xz\right\}$$
So, $$LHS=(x+y+z)\left\{x^2+y^2+z^2-xy-yz-xz\right\}$$
So, $$LHS=RHS$$.
Hence it is proved that $x^3+y^3+z^3-3xyz=(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$.

Note: Without using the identity we can also prove $x^3+y^3+z^3-3xyz=(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$. For this method take first $$RHS=(x+y+z)\left\{x^2+y^2+z^2-xy-yz-xz\right\}$$. After this use multiplication of polynomials to simplify the equation.
So, $$RHS=x{x}^2+x{y}^2+x{z}^2-xxy-xyz-xzx+y{x}^2+y{y}^2+y{z}^2-yxy-yyz-yxz+z{x}^2+z{y}^2+z{z}^2-zxy-zyz-zxz$$
Simplifying the equation,
$$RHS=x^3+x{y}^2+x{z}^2-x^{2}y-xyz-x^{2}z+y{x}^2+y^3+y{z}^2-x{y}^2-y^{2}z-xyz+z{x}^2+z{y}^2+z^3-xyz-y{z}^2-x{z}^2$$
Arranging the terms, $$RHS=(x^3+y^3+z^3)+(x{y}^2+x{z}^2-x^{2}y-x^{2}z+x^{2}y+y{z}^2-x{y}^2-y^{2}z+x^{2}z+z{y}^2-y{z}^2-x{z}^2)-3xyz$$
Simplifying, $$RHS=(x^3+y^3+z^3)+(0)-3xyz$$
So, $$RHS=x^3+y^3+z^3-3xyz$$
So, $$RHS=LHS$$.
Hence it is proved that $x^3+y^3+z^3-3xyz=(x+y+z)\cdot (x^2+y^2+z^2-xy-yz-zx)$