Answer
398.4k+ views
Hint: In this problem, we have to find the value of the given cubic equation for the given condition. We should know that to solve these types of problems we should know some algebraic formulas such as cubic formulas. We can take the equation \[x+y=5\] and take a cube on both sides, we can then expand the whole cubic using an algebraic cubic formula to find the value of the given equation.
Complete step by step answer:
We know that the given equation to which we have to find the value is,
\[{{x}^{3}}+{{y}^{3}}+15xy-125\]
We also know that the given condition for the above equation is,
\[x+y=5\] ….. (1)
We can take cube on both the sides of the equation (1), we get
\[\begin{align}
& \Rightarrow {{\left( x+y \right)}^{3}}={{5}^{3}} \\
& \Rightarrow {{\left( x+y \right)}^{3}}=125 \\
\end{align}\]
We know that the algebraic cubic formula is,
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}\]
Now we can apply this formula to expand the above step, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}=125\]
Now we can take the common terms, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)=125\]
We can now substitute the given condition (1), in the above step, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3xy\left( 5 \right)=125\]
Now we can subtract 125 on both sides, we get
\[\begin{align}
& \Rightarrow {{x}^{3}}+{{y}^{3}}+15xy-125=125-125 \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+15xy-125=0 \\
\end{align}\]
Therefore, the value of \[{{x}^{3}}+{{y}^{3}}+15xy-125\] is 0.
Note: Students make mistakes while writing the correct algebraic formulas. We should know that to solve these types of problems, we should know some algebraic formulas, properties and identities. We should also remember some cube terms to substitute in the problem, when required. We should also know some simplification method, to get the required result.
Complete step by step answer:
We know that the given equation to which we have to find the value is,
\[{{x}^{3}}+{{y}^{3}}+15xy-125\]
We also know that the given condition for the above equation is,
\[x+y=5\] ….. (1)
We can take cube on both the sides of the equation (1), we get
\[\begin{align}
& \Rightarrow {{\left( x+y \right)}^{3}}={{5}^{3}} \\
& \Rightarrow {{\left( x+y \right)}^{3}}=125 \\
\end{align}\]
We know that the algebraic cubic formula is,
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}\]
Now we can apply this formula to expand the above step, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}=125\]
Now we can take the common terms, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)=125\]
We can now substitute the given condition (1), in the above step, we get
\[\Rightarrow {{x}^{3}}+{{y}^{3}}+3xy\left( 5 \right)=125\]
Now we can subtract 125 on both sides, we get
\[\begin{align}
& \Rightarrow {{x}^{3}}+{{y}^{3}}+15xy-125=125-125 \\
& \Rightarrow {{x}^{3}}+{{y}^{3}}+15xy-125=0 \\
\end{align}\]
Therefore, the value of \[{{x}^{3}}+{{y}^{3}}+15xy-125\] is 0.
Note: Students make mistakes while writing the correct algebraic formulas. We should know that to solve these types of problems, we should know some algebraic formulas, properties and identities. We should also remember some cube terms to substitute in the problem, when required. We should also know some simplification method, to get the required result.
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