Question

hey guys please solve that...​

Answer 1

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\huge{\pink{\boxed{\green{\underline{\red{\sf{TO\:PROOF:}}}}}}}$

$\huge{\red{\boxed{\pink{\sf{b^{3}+a^{2}c+ac^{2}=3abc}}}}}$

Let one of the root of this given eqn

$\to first \: zeroes \: (\alpha ) \\ \to \: second \: zeroes \: ({ \alpha })^{2} \\ {\boxed{given \: eqn}} \\ \to \: a {x}^{2} + bx + c = 0$

$\therefore \: sum \: of \: roots( \alpha + { \alpha }^{2} ) = \frac{ - b}{a} - - - - - - (1) \\ \therefore \: product \: of \: roots( \alpha \times { \alpha }^{2} ) = \frac{c}{a} \to \: { \alpha }^{3} = \frac{c}{a} - - - - - (2)$

Taking cube of (1)

$\to( { \alpha + { \alpha }^{2} })^{3} = \frac{ - {b}^{3} }{ {a}^{3} } \\ \to{ \alpha }^{3} + \alpha^{6} + 3 \times \alpha \times { \alpha }^{4} + 3 { \alpha }^{2} \times { \alpha }^{2} = \frac{ - b^{3} }{ {a}^{3} } \\ \to \: { \alpha }^{3} + {( { \alpha }^{3} })^{2} + 3 \alpha ^{3} ( { \alpha }^{2} + \alpha ) = \frac{ - {b}^{3} }{ {a}^{3} } - - - - - - (3)$

Substituting the value from (1) and (2)

$\to \: \frac{c}{a} + (\frac{c}{a} ) {}^{2} + 3 \times \frac{c}{a} ( \frac{ - b}{a} ) = \frac{ - {b}^{3} }{ {a}^{2} } \\ \to \: \frac{c}{a} + \frac{ {c}^{2} }{ {a}^{2} } - \frac{3bc}{ {a}^{2} } = \frac{ - {b}^{3} }{ {a}^{3} }$

Multiplying the whole eqn by a^3

$\to \: {a}^{2} c + a {c}^{2} - 3abc = - {b}^{3} \\ \to \: {b}^{2} + {a}^{2} c + {ac}^{2} = 3abc$

$\huge{\red{\boxed{\green{\underline{\sf{PROVED}}}}}}$

Answer 2

Step-by-step explanation:

→firstzeroes(α)→secondzeroes(α)2giveneqn→ax2+bx+c=0

\begin{lgathered}\therefore \: sum \: of \: roots( \alpha + { \alpha }^{2} ) = \frac{ - b}{a} - - - - - - (1) \\ \therefore \: product \: of \: roots( \alpha \times { \alpha }^{2} ) = \frac{c}{a} \to \: { \alpha }^{3} = \frac{c}{a} - - - - - (2) \\\end{lgathered}∴sumofroots(α+α2)=a−b−−−−−−(1)∴productofroots(α×α2)=ac→α3=ac−−−−−(2)

Taking cube of (1)

\begin{lgathered}\to( { \alpha + { \alpha }^{2} })^{3} = \frac{ - {b}^{3} }{ {a}^{3} } \\ \to{ \alpha }^{3} + \alpha^{6} + 3 \times \alpha \times { \alpha }^{4} + 3 { \alpha }^{2} \times { \alpha }^{2} = \frac{ - b^{3} }{ {a}^{3} } \\ \to \: { \alpha }^{3} + {( { \alpha }^{3} })^{2} + 3 \alpha ^{3} ( { \alpha }^{2} + \alpha ) = \frac{ - {b}^{3} }{ {a}^{3} } - - - - - - (3)\end{lgathered}→(α+α2)3=a3−b3→α3+α6+3×α×α4+3α2×α2=a3−b3→α3+(α3)2+3α3(α2+α)=a3−b3−−−−−−(3)

Substituting the value from (1) and(2)

\begin{lgathered}\to \: \frac{c}{a} + (\frac{c}{a} ) {}^{2} + 3 \times \frac{c}{a} ( \frac{ - b}{a} ) = \frac{ - {b}^{3} }{ {a}^{2} } \\ \to \: \frac{c}{a} + \frac{ {c}^{2} }{ {a}^{2} } - \frac{3bc}{ {a}^{2} } = \frac{ - {b}^{3} }{ {a}^{3} }\end{lgathered}→ac+(ac)2+3×ac(a−b)=a2−b3→ac+a2c2−a23bc=a3−b3

Multiplying the whole eqn by a^3

\begin{lgathered}\to \: {a}^{2} c + a {c}^{2} - 3abc = - {b}^{3} \\ \to \: {b}^{2} + {a}^{2} c + {ac}^{2} ={PROVED