Math, asked by avinashsingh48, 1 year ago

if a + 1/5a = 1 then evaluate a³ + 1/125×a³ ?​

Answers

Answered by Anonymous
3

Step-by-step explanation:

Given :-

This relation is given

a + \frac{1}{5a} = 1

To find:

\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } \\\end{lgathered}

Solve as follows:

By using identity

\begin{lgathered}{(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a+b) \\ \\\end{lgathered}

Put in the values:

\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } = {(a + \frac{1}{5a} )}^{3} - 3a\frac{1}{5a}(a + \frac{1}{5a} ) \\ \\\end{lgathered}

Put the value from relation:

\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } = {1}^{3} - 3\frac{1}{5}(1) \\ \\ = 1 + \frac{3 }{5}\\ \\ = - \frac{2}{5}\end{lgathered}

Answered by itzyourdeathgirl
7

Answer:

Given :-

This relation is given

a + \frac{1}{5a} = 1a+

5a

1

=1

To find:

\begin{gathered}\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } \\\end{lgathered} \end{gathered}

Solve as follows:

By using identity

\begin{gathered}\begin{lgathered}{(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a+b) \\ \\\end{lgathered} \end{gathered}

Put in the values:

\begin{gathered}\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } = {(a + \frac{1}{5a} )}^{3} - 3a\frac{1}{5a}(a + \frac{1}{5a} ) \\ \\\end{lgathered} \end{gathered}

Put the value from relation:

\begin{gathered}\begin{lgathered}{a}^{3} + \frac{1}{125 {a}^{3} } = {1}^{3} - 3\frac{1}{5}(1) \\ \\ = 1 + \frac{3 }{5}\\ \\ = - \frac{2}{5}\end{lgathered} \end{gathered}

i hope it's help you ❤

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