Question

Factorize the following equation:
${x^3} + {y^3} + {z^3} = 3xyz$

Answer 1

Hint: To factorize we convert the given equation such that the RHS is equal to zero and express it as a product of at least two terms. We try to express it as a product of (x + y + z) and another term.

Complete step-by-step answer:
Given Data,
${x^3} + {y^3} + {z^3} = 3xyz$
This equation can also be written as
$ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 0$
Now, we will add and subtract the terms $3{x^2}y$ and $3x{y^2}$ to this equation.
\[ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz + 3{x^2}y - 3{x^2}y + 3x{y^2} - 3x{y^2} = 0\]
We know the formula for a term ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$, comparing this with the above equation we get a = x and b = y.
$ \Rightarrow {\left( {x + y} \right)^3} + {z^3} - 3xy\left( {x + y + z} \right) = 0$
Now we take (x + y + z) common out of the equation, we add and subtract a few terms to simplify the equation.
$ \Rightarrow \left( {x + y + z} \right)\left[ {{{\left( {x + y} \right)}^2} + {z^2} - \left( {x + y} \right)z} \right] - 3xy\left( {x + y + z} \right) = 0$
\[ \Rightarrow \left( {x + y + z} \right)\left[ {{x^2} + 2xy + {y^2} + {z^2} - yz - xz - 3xy} \right] = 0\]
\[ \Rightarrow \left( {x + y + z} \right)\left[ {{x^2} + {y^2} + {z^2} - yz - xz - xy} \right] = 0\]
Hence factorized.

Note: In order to solve this type of problems the key is to know how to factorize a cubic equation. The key step is to add and subtract terms to make it in the form of an equation for which we can apply known formulae. We used these formulae in factorizing this equation, ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$and ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$