
Answers
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 ✌️