Question

If \[x + y + z = 8\] and \[xy + yz + zx = 20\] find value of \[{x^3} + {y^3} + {z^3} - 3xyz\].

Answer 1

Hint: We need to use the algebraic identities and expansion here, because a partial problem is a part of an algebraic identity.
Use formula :
\[
  {(x + y + z)^2} = \left( {x + y + z} \right)\left( {x + y + z} \right) \\
   \Rightarrow xx + xy + xz + yx + yy + yz + zx + zy + zz \\
   \Rightarrow {x^2} + 2xy + 2xz + 2yz + {y^2} + {z^2} \\
   \Rightarrow {x^2} + {y^2} + {z^2} + 2(xy + yz + zx) \\
\]
\[{x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)\]

Complete step-by-step answer:
Given that,
\[x + y + z = 8\]
\[xy + yz + zx = 20\]
$\Rightarrow$ \[{x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)\]
\[{x^3} + {y^3} + {z^3} - 3xyz= (x + y + z)({x^2} + {y^2} + {z^2} + 2(xy + yz + zx) - 3(xy + yz + zx))\]
$\Rightarrow$ \[{x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({(x + y + z)^2} - 3(xy + yz + zx))\]
Now substitute the given value in right side of equation to get the answer
$(x + y + z)({(x + y + z)^2} - 3(xy + yz + zx)$
=$8({8^2} - 3(20))$
=$8(64 - 60)$
=$8 \times 4$
=32
Hence 32 is the correct answer.
Additional information:
There are many algebraic identities in mathematics that makes the calculative part easy.
These satisfy their equation for any value of the variable.
There are some identities that use binomial theorems for their expansion. These are frequently used identities. They are available upto power 4 but we can extend them as we need.
The factoring formulas used are also the algebraic identities. These are used when we need to split a number and then use its expansion.
There exist some three variable identities. We have used it in our solution.

Note: Students just have to use the algebraic identities.
Sometimes they need to get modified or rearranged according to the requirement.