Question

If \[x + y + 2 = 0\] then find the value of \[{x^3} + {y^3} + {\text{8}}\]

Answer 1

Hint: We have many mathematical functions having it’s expression defined and their behaviour too in some special cases or else cube both the side and reduce RHS to obtain optimize value.

Complete step-by-step answer:
Let’s begin with what is given it is said that \[x + y + 2 = 0\] and to find \[{x^3} + {y^3} + {\text{8}}\], we can imagine 2 is 3rd variable z so \[x + y + z = 0\] and we are asked for \[{x^3} + {y^3} + {z^3} = {\text{?}}\]
So we know that
\[{x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right){\text{ }}\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)\]
If we can imagine the given value 2 as z
We get
\[{x^3} + {y^3} + 8 - 3xy \times 2 = \left( {x + y + z} \right){\text{ }}\left( {{x^2} + {y^2} + 4 - xy - 2y - 2x} \right)\]
as we know that \[x + y + z = 0\], the RHS will we
\[
  {x^3} + {y^3} + {\text{8}} - 6xy = \left( 0 \right)\left( {{x^2} + {y^2} + 4 - xy - 2y - 2x} \right) \\
  {x^3} + {y^3} + 8 = 6xy \\
\]
If we don’t get this formula, we can derive it by cubing both the side like
\[
  {\left( {x + y + 2} \right)^3} = {\text{0}} \\
   = {\left( {x + y} \right)^3} + {2^3} + 3{\text{ }}\left( {x + y} \right)\left( 2 \right)\left( {x + y + 2} \right) \\
\]
as we know \[x + y + 2 = 0\] then putting value we get
\[
   = {\left( {x + y} \right)^3} + 8 + 3\left( {x + y} \right)\left( 2 \right)\left( 0 \right) \\
   = {x^3} + {y^3}{\text{ + }}3xy\left( {x + y} \right) + 8 \\
  0 = {x^3} + {y^3} + 8 + 3xy\left( {x + y} \right) \\
\]
as we know
\[
  x + y + 2 = 0 \\
  {\text{so x}} + y = - 2 \\
  {\text{replacing }}x + y{\text{ as}} - 2,{\text{we get}} \\
  0 = {x^3} + {y^3} + 8 + 3xy\left( { - 2} \right) \\
  {\text{Hence}},{\text{ }}{x^3} + {y^3} + 8 = 6xy \\
\]
So value of ${x^3} + {y^3} + 8$ is 6xy.

Note: The direct function derived should be frequent and fast to analyse the situation to choose appropriate one or else everything can be derived from the root, but it’s lengthy and inconvenient to use each and every time.