Question

# Find the value of $8{{x}^{3}}+27{{y}^{3}}$ if $2x+3y=13$ and $xy=6$.

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Hint: To find the value of the given algebraic expression, use the algebraic identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$. Substitute $a=2x,b=3y$ in the algebraic identity and simplify the equation using the equations given in the question to get the value of $8{{x}^{3}}+27{{y}^{3}}$.

We know that for two variables ‘x’ and ‘y’, we have $2x+3y=13$ and $xy=6$. We have to calculate the value of the algebraic expression $8{{x}^{3}}+27{{y}^{3}}$.
To do so, we will use the algebraic identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Substituting $a=2x,b=3y$ in the above algebraic identity, we have ${{\left( 2x+3y \right)}^{3}}={{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+3\left( 2x \right)\left( 3y \right)+\left( 2x+3y \right)$.
Simplifying the above equation, we have ${{\left( 2x+3y \right)}^{3}}={{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+3\left( 2x \right)\left( 3y \right)+\left( 2x+3y \right)=8{{x}^{3}}+27{{y}^{3}}+18xy\left( 2x+3y \right)$.
Substituting $2x+3y=13$ and $xy=6$ in the above equation ${{\left( 13 \right)}^{3}}=8{{x}^{3}}+27{{y}^{3}}+18\times 6\times 13$.
Simplifying the above equation, we have $17069=8{{x}^{3}}+27{{y}^{3}}+1404$.
Rearranging the terms of the above equation, we have $8{{x}^{3}}+27{{y}^{3}}=17069-1404=15665$.
Hence, the value of the expression $8{{x}^{3}}+27{{y}^{3}}$ is 15665.
Note: We can also calculate the value of given algebraic expression by solving the linear equations $2x+3y=13$ and $xy=6$ by elimination method and finding the values of variables ‘x’ and ‘y’. then substitute those values in the given algebraic expression and calculate its value. Another way to solve this question is by using the algebraic identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$.