Math, asked by ipsita1301, 5 months ago


  \\ if \: \frac{a}{b}  +   \frac{b}{a}  =  - 1 \: then  \:  {a}^{3}  -  {b}^{3}  =    {?}


Answers

Answered by Anonymous
30

 \huge \green{given : } \:  \\  \:  \:  \:  \:  \:  \:  \frac{a}{b}  +  \frac{b}{a}  =  - 1

 \huge \green{to \: find : } \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  {a}^{3}  -  {b}^{3}  =  {?}

 \huge  \green{solution : }

 \sf \:  \:  \:  \:  \implies \:  \frac{a}{b}  +  \frac{ b}{a}  =  - 1 \\

 \sf \:  \:  \:  \:  \:  \implies \frac{ { a}^{2} +  {b}^{2}  }{ba}  =  - 1 \\

 \sf \:  \:  \:  \:  \:   \:  \boxed{ \implies {a}^{2}  +  {b}^{2} =  - ab }

 \huge \red{Equation : }

 \bf \: ( {a}^{3}  -  {b}^{3} ) = \blue{ (a -  b)( {a}^{2}  + ab +  {b}^{2} )}

 \bf \:  \:  \: \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \blue{ = (a - b)( {a}^{2} +  {b}^{2}   + ab)}

 \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = \blue{(a - b)( -  \cancel{ab} +  \cancel{ab})}

 \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \blue{ = (a - b) - 1}

 \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \blue{ = 0}

 \green{Hence \:  proved }

 \:  \:  \sf \boxed{ {a}^{3}  -  {b}^{3} =  - 1 }

Answered by swanandi01
6

Answer:

given:

b

a

+

a

b

=−1

\begin{gathered} \huge \green{to \: find : } \\ \: \: \: \: \: \: \: \: \: {a}^{3} - {b}^{3} = {?}\end{gathered}

tofind:

a

3

−b

3

=?

\huge \green{solution : }solution:

\begin{gathered} \sf \: \: \: \: \implies \: \frac{a}{b} + \frac{ b}{a} = - 1 \\ \end{gathered}

b

a

+

a

b

=−1

\begin{gathered} \sf \: \: \: \: \: \implies \frac{ { a}^{2} + {b}^{2} }{ba} = - 1 \\ \end{gathered}

ba

a

2

+b

2

=−1

\sf \: \: \: \: \: \: \boxed{ \implies {a}^{2} + {b}^{2} = - ab }

⟹a

2

+b

2

=−ab

\huge \red{Equation : }Equation:

\bf \: ( {a}^{3} - {b}^{3} ) = \blue{ (a - b)( {a}^{2} + ab + {b}^{2} )}(a

3

−b

3

)=(a−b)(a

2

+ab+b

2

)

\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ = (a - b)( {a}^{2} + {b}^{2} + ab)}=(a−b)(a

2

+b

2

+ab)

\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \blue{(a - b)( - \cancel{ab} + \cancel{ab})}=(a−b)(−

ab

+

ab

)

\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ = (a - b) - 1}=(a−b)−1

\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ = 0}=0

\green{Hence \: proved }Henceproved

\: \: \sf \boxed{ {a}^{3} - {b}^{3} = - 1 }

a

3

−b

3

=−1

hope it helps you ✌️

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