Math, asked by avinashsingh48, 11 months ago

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

Answers

Answered by abdul9838
1

 <b> <body \: bgcolor  = "skyblue">

 \small\bf \red{hey \: user \: here \: is \: ans} \\  \\  \huge \bf \blue{solution} \\  \\  \small \bf \pink{given \: that} \\  \\ \small \bf \pink{a +  \frac{1}{5a}  = 1} \\  \\ \small \bf \pink{we \: have \: to \: find \: the \: value \: of \: } \\ \small \bf \pink{ {a}^{3}  +  \frac{1}{125 {a}^{3} } } \\  \\ \small \bf \pink{let \: start} \\  \\ \small \bf \pink{as \: given} \\  \\ \small \bf \pink{a +  \frac{1}{5a}  = 1 \:  \:  \: equation \: 1st}  \\  \\ \small \bf \blue{ \underline{cubing \: on \: both \: sides}} \\  \\ \small \bf \pink{ {(a +  \frac{1}{5a} )}^{3} =  {1}^{3}  } \\  \\ \small \bf \red{ \huge \: now} \\  \\ \small \bf \orange{using \: to \: this \: identity} \\  \\ \small \bf \green{(a + b)^{3}  =  {a}^{3}  +  {b}^{3} + 3ab(a + b) } \\  \\ \small \bf \pink{ {a}^{3}  +  \frac{1}{125 {a}^{3} }  + 3 \times a \times  \frac{1}{5a}(a  +  \frac{1}{5a}  ) = 1 } \\  \\ \small \bf \pink{ {a}^{3} +  \frac{1}{125 {a}^{3} } + 3 \times  \cancel{a}  \times  \frac{1}{5 \cancel{a}} \times  ( 1) = 1} \\ \small \bf \pink{hence \: a +  \frac{1}{5a}  = 1(from \: equation \: 1st )} \\  \\ \small \bf \pink{ {a}^{3}  +  \frac{1}{125 {a}^{3} } +  \frac{3}{5}   = 1} \\  \\ \small \bf \pink{ {a}^{3} +  \frac{1}{125 {a}^{3} }  = 1 -  \frac{3}{5}  } \\  \\ \small \bf \pink{ {a}^{ 3} +  \frac{1}{125 {a}^{3} }  =  \frac{5 - 3}{5}  } \\  \\ \small \bf \pink{ {a}^{3}  +  \frac{1}{125 {a}^{3} } =  \frac{2}{5}  \:  \: answer }

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 \bf \boxed{ \bf \blue {\: thanks \: for \: asking}} \\  \bf \purple{abdul \: is \: here} \\  \bf \green{ \huge \: follow \: me}

Answered by itzyourdeathgirl
10

Answer:

b><bodybgcolor="skyblue">

\begin{gathered} \small\bf \red{hey \: user \: here \: is \: ans} \\ \\ \huge \bf \blue{solution} \\ \\ \small \bf \pink{given \: that} \\ \\ \small \bf \pink{a + \frac{1}{5a} = 1} \\ \\ \small \bf \pink{we \: have \: to \: find \: the \: value \: of \: } \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } } \\ \\ \small \bf \pink{let \: start} \\ \\ \small \bf \pink{as \: given} \\ \\ \small \bf \pink{a + \frac{1}{5a} = 1 \: \: \: equation \: 1st} \\ \\ \small \bf \blue{ \underline{cubing \: on \: both \: sides}} \\ \\ \small \bf \pink{ {(a + \frac{1}{5a} )}^{3} = {1}^{3} } \\ \\ \small \bf \red{ \huge \: now} \\ \\ \small \bf \orange{using \: to \: this \: identity} \\ \\ \small \bf \green{(a + b)^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) } \\ \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } + 3 \times a \times \frac{1}{5a}(a + \frac{1}{5a} ) = 1 } \\ \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } + 3 \times \cancel{a} \times \frac{1}{5 \cancel{a}} \times ( 1) = 1} \\ \small \bf \pink{hence \: a + \frac{1}{5a} = 1(from \: equation \: 1st )} \\ \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } + \frac{3}{5} = 1} \\ \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } = 1 - \frac{3}{5} } \\ \\ \small \bf \pink{ {a}^{ 3} + \frac{1}{125 {a}^{3} } = \frac{5 - 3}{5} } \\ \\ \small \bf \pink{ {a}^{3} + \frac{1}{125 {a}^{3} } = \frac{2}{5} \: \: answer }\end{gathered}

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giventhat

< marquee \: behavior = "alternate" > < marquee \: bgcolor = "red" > <marqueebehavior="alternate"><marqueebgcolor="red">

\begin{gathered} \bf \boxed{ \bf \blue {\: thanks \: for \: asking}} \\ \bf \purple{abdul \: is \: here} \\ \bf \green{ \huge \: follow \: me}\end{gathered}

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