Question

Pls solve this
options are
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Answer 1

Answer:

00⁰00000000000⁰0000⁰0000⁰0000⁰000000000⁰00000⁰0000000

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a + b + c = 0$

On squaring both sides, we get

$\rm :\longmapsto\: {(a + b + c)}^{2} = 0$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 0$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = - 2(ab + bc + ca )$

On squaring both sides, we get

$\rm :\longmapsto\: ({a}^{2} + {b}^{2} + {c}^{2} )^{2} = 4(ab + bc + ca )^{2}$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: 4\bigg( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} + 2(ab)(bc) + 2(bc)(ca) + 2(ca)(ab)\bigg)$

$\boxed{ \sf{ \: \because \: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}}$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: 4\bigg( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} + 2( {b}^{2} ac) + 2( {c}^{2} ba) + 2( {a}^{2} cb)\bigg)$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} = \: \: \: \: \: \: \: \: \:\: \\ \rm \: 4\bigg( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} + 2abc(a + b + c)\bigg)$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} = \\ \rm \: 4\bigg( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} + 2abc(0)\bigg) \: \: \: \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \: \because \: a + b + c = 0}}$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} = \: \: \: \: \: \: \: \: \:\: \\ \rm \: 4\bigg( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} \bigg)$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2} =\\ \rm \: 4 {a}^{2} {b}^{2} +4{b}^{2} {c}^{2} + 4{c}^{2} {a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} = 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2}$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} = 2 ({a}^{2} {b}^{2} +{b}^{2} {c}^{2} +{c}^{2} {a}^{2})$

$\bf\implies \: \dfrac{ {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} }{ {a}^{4} + {b}^{4} + {c}^{4} } = \dfrac{1}{2}$

Additional Information :-

$\boxed{ \sf{ \:(a + b)^{2} = {a}^{2} + 2ab + {b}^{2}}}$

$\boxed{ \sf{ \:(a - b)^{2} = {a}^{2} - 2ab + {b}^{2}}}$

$\boxed{ \sf{ \:(a + b)^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)}}$

$\boxed{ \sf{ \:(a - b)^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)}}$

$\boxed{ \sf{ \: {(a + b)}^{2} = {(a - b)}^{2} + 4ab}}$

$\boxed{ \sf{ \: {(a - b)}^{2} = {(a + b)}^{2} - 4ab}}$