Question

$\green{\boxed{\sf \pink {if \: \frac{a}{b} + \frac{b}{a} = 1 \: \: \: \: \: \: then \: a^3 +b^3=}}}$

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Answer 1

Answer:

Hope it helps you brother.

Answer 2

${\large{\bf{\underbrace{\underline{Question:-}}}}}$

Q)) If $\sf\dfrac{a}{b} +\dfrac{b}{a} = 1$ Then, find : a³ + b³

${\large{\bf{\underbrace{\underline{Answer\checkmark}}}}}$

We have :

• $\sf\dfrac{a}{b}+\dfrac{b}{c}=1$

${\textit{\textbf{\underline{Modifying\:it:}}}}$

$\colon \implies \sf \: \frac{a}{b} + \frac{b}{a} = 1 \\ \\ \colon \implies \: \sf \frac{a \times a + b \times b}{ab} = 1 \\ \\ \colon \implies \sf \: \frac{ {a}^{2} + {b}^{2} }{ab} = 1 \\ \\ \colon \implies \boxed{ \tt \blue{ {a}^{2} + {b}^{2} = ab}} \: .... \sf(i)$

${\underline{\textit{\textbf{Finding\:(a+b):}}}}$

$\mapsto \sf \: {(a + b)}^{2} = \underline{{a}^{2} + {b}^{2}} + 2ab \\ \\ \sf \: from \: eq(i) \\ \\ \mapsto \sf \: {(a + b)}^{2} = ab + 2ab \\ \\ \mapsto \sf \: {(a + b)}^{2} = 3ab \\ \\ \longrightarrow \: \boxed{ \purple{ \frak{(a + b) = \sqrt{3ab}}}} \: \sf ....(ii)$

Now,

${\underline{\textit{\textbf{Finding\:Final\:Answer:}}}}$

$\colon \rightarrow \: \sf {a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab) \\ \\ \sf \: from \: eq(i) \: and \: eq(ii) \\ \\ \colon \rightarrow \sf \: {a}^{3} + {b}^{3} = \sqrt{3ab} (ab - ab) \\ \\ \colon \rightarrow \sf \: {a}^{3} + {b}^{3} = \sqrt{3ab} \times 0 \\ \\ \star \longrightarrow \: \underline{ \boxed{ \tt{ \pink{ {a}^{3} } + \red{{b}^{3}} = \green{ \bf 0}}}}$

So ,

The value of a³ + b³ would be Zero (0) .

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${\bf{\underline{\underline{Some~Important~Formulas:}}}}$

${ \boxed{ \boxed{ \begin{array}{c} \pink{ \tt \bull \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy } \\ \\ \purple{\bull \tt \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy} \\ \\ \pink{ \bull \: \tt {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)} \\ \\ \purple{ \bull \tt \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)} \\ \\ \pink{ \tt \bull \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy) } \\ \\ \purple{\tt \bull \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy) }\end{array}}}}$