Find the value of given expression $\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$ by reducing
it into the simplified form.
\[\begin{array}{*{35}{l}}
A)\text{ }445.99 \\
B)\text{ }500 \\
C)\text{ }400 \\
D)\text{ }600 \\
\end{array}\]
Answer
618.6k+ views
Hint:We know that the algebraic expression ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)({{x}^{2}}- xy+{{y}^{2}})$.Therefore consider $x=360\And y=139$ and substitute in the formula.And reduce the expression to get the actual value or solve it directly with those values by expanding the squares and
cubes.
Complete step by step answer:
We have given that $\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
We know that that the ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)({{x}^{2}}-xy+{{y}^{2}})$⟶equation(1)
Let us consider,
$x=360$
$y=139$
By shifting $\left( x+y \right)$ term to the L.H.S and ${{x}^{3}}+{{y}^{3}}$term to the R.H.S in equation(1) we get
⟹$\left( x+y \right)=\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$⟶equation(2)
The given expression in problem is as exactly equal to that of R.H.S of equation(2)
That is , $\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$=$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
Now L.H.S = $\left( x+y \right)$=$(360+139)$=500
Therfore, R.H.S=L.H.S
That is, $\left( x+y \right)=\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$
$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$=$(360+139)$=500
Therefore the correct option is (B)
Note:
We can also solve this by expanding the given expression step by step each term in the problem as follows.
Alternative Method:
⟹$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
⟹$\dfrac{46656000+2685619}{130321-50179+19321}$
On simplifying we get
⟹$496.0801\cong 500$
cubes.
Complete step by step answer:
We have given that $\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
We know that that the ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)({{x}^{2}}-xy+{{y}^{2}})$⟶equation(1)
Let us consider,
$x=360$
$y=139$
By shifting $\left( x+y \right)$ term to the L.H.S and ${{x}^{3}}+{{y}^{3}}$term to the R.H.S in equation(1) we get
⟹$\left( x+y \right)=\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$⟶equation(2)
The given expression in problem is as exactly equal to that of R.H.S of equation(2)
That is , $\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$=$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
Now L.H.S = $\left( x+y \right)$=$(360+139)$=500
Therfore, R.H.S=L.H.S
That is, $\left( x+y \right)=\dfrac{{{x}^{3}}+{{y}^{3}}}{({{x}^{2}}-xy+{{y}^{2}})}$
$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$=$(360+139)$=500
Therefore the correct option is (B)
Note:
We can also solve this by expanding the given expression step by step each term in the problem as follows.
Alternative Method:
⟹$\dfrac{{{\left( 360 \right)}^{3}}+{{\left( 139 \right)}^{3}}}{{{\left( 361 \right)}^{2}}-361\times 139+{{\left( 139 \right)}^{2}}}$
⟹$\dfrac{46656000+2685619}{130321-50179+19321}$
On simplifying we get
⟹$496.0801\cong 500$
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