
A: Find the values of each expression:
(i) ${{2}^{3}}+{{1}^{3}}$
(ii) $\sqrt{{{1}^{3}}+{{2}^{3}}}$
B. Find the value of $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}$.
C. Find the value of $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}}$.
D. Find the value of $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}}$.
Answer
418.2k+ views
Hint: For solving this type of questions you should know about general mathematics calculations. In these problems, we will simply just find the values of the cubes of the given numbers and then add them and then we will take their square roots and solve them.
Complete step by step answer:
According to our question, we have to find the values of some of the given expressions. For solving these expressions, first we will find the values of the given terms as cube form. And then we will find the submission as asked in the question and then take the square root of that number. So, let us solve each of them one by one.
A. (i) ${{2}^{3}}+{{1}^{3}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
\end{align}$
So, we will substitute these values in the expression and get,
$=1+8=9$
(ii) $\sqrt{{{1}^{3}}+{{2}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
\end{align}$
So, we will substitute these values in the expression and get,
$=\sqrt{1+8}=\sqrt{9}=3$
B. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
& {{3}^{3}}=3\times 3\times 3=27 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27}=\sqrt{1+35}=\sqrt{36}=6$
C. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1 \\
& {{2}^{3}}=8 \\
& {{3}^{3}}=27 \\
& {{4}^{3}}=4\times 4\times 4=64 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27+64}=\sqrt{36+64}=\sqrt{100}=10$
D. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1 \\
& {{2}^{3}}=8 \\
& {{3}^{3}}=27 \\
& {{4}^{3}}=64 \\
& {{5}^{3}}=5\times 5\times 5=125 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27+64+125}=\sqrt{100+125}=\sqrt{225}=15$
So, these are the values that we got.
Note: While solving such questions you have to be careful about the calculations because there are not many methods for solving them. But the only important thing is the calculations, so take the cubes accurately and then take the square root carefully, otherwise the whole solution will be wrong.
Complete step by step answer:
According to our question, we have to find the values of some of the given expressions. For solving these expressions, first we will find the values of the given terms as cube form. And then we will find the submission as asked in the question and then take the square root of that number. So, let us solve each of them one by one.
A. (i) ${{2}^{3}}+{{1}^{3}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
\end{align}$
So, we will substitute these values in the expression and get,
$=1+8=9$
(ii) $\sqrt{{{1}^{3}}+{{2}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
\end{align}$
So, we will substitute these values in the expression and get,
$=\sqrt{1+8}=\sqrt{9}=3$
B. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1\times 1\times 1=1 \\
& {{2}^{3}}=2\times 2\times 2=8 \\
& {{3}^{3}}=3\times 3\times 3=27 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27}=\sqrt{1+35}=\sqrt{36}=6$
C. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1 \\
& {{2}^{3}}=8 \\
& {{3}^{3}}=27 \\
& {{4}^{3}}=4\times 4\times 4=64 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27+64}=\sqrt{36+64}=\sqrt{100}=10$
D. $\sqrt{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}}$
As we know that,
$\begin{align}
& {{1}^{3}}=1 \\
& {{2}^{3}}=8 \\
& {{3}^{3}}=27 \\
& {{4}^{3}}=64 \\
& {{5}^{3}}=5\times 5\times 5=125 \\
\end{align}$
So, we will substitute these values in the given expression and so we get,
$=\sqrt{1+8+27+64+125}=\sqrt{100+125}=\sqrt{225}=15$
So, these are the values that we got.
Note: While solving such questions you have to be careful about the calculations because there are not many methods for solving them. But the only important thing is the calculations, so take the cubes accurately and then take the square root carefully, otherwise the whole solution will be wrong.
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