Question

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Question:-

$a + b + c = 4$

${a}^{2} + {b}^{2} + {c}^{2} = 10$

${a}^{3} + {b}^{3} + {c}^{3} = 22$

${a}^{4} + {b}^{4} + {c}^{4} = \: ?$

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Answer 1

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

Given that

$\rm :\longmapsto\:a + b + c = 4$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 10$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + {c}^{3} = 22$

Now, Consider

$\rm :\longmapsto\:a + b + c = 4$

On squaring both sides, we get

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

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 16$

$\rm :\longmapsto\: 10 + 2(ab + bc + ca) = 16$

$\rm :\longmapsto\: 2(ab + bc + ca) = 16 - 10$

$\rm :\longmapsto\: 2(ab + bc + ca) = 6$

$\rm \implies\:\boxed{\tt{ ab + bc + ca = 3}} - - - (1)$

Now, Consider

$\rm :\longmapsto\:(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2}) = 4 \times 10$

$\rm :\longmapsto\: {a}^{3} + {ab}^{2} + {ac}^{2} +{ba}^{2} +{b}^{3} + {bc}^{2} + {ca}^{2} + {cb}^{2} + {c}^{3} = 40$

$\rm :\longmapsto\: {a}^{3}+ {b}^{3} + {c}^{3} + {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40$

$\rm :\longmapsto\: 22+ {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40$

$\red{ \bigg\{ \sf \: \because \: of \: given \bigg\}}$

$\rm :\longmapsto\:{ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40 - 22$

$\rm :\longmapsto\:\boxed{\tt{ {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=18}} - - - (2)$

Consider

$\rm :\longmapsto\:(a + b + c)(ab + bc + ca) = 4 \times 3$

$\rm :\longmapsto\: {ba}^{2} +abc + {ca}^{2} + {ab}^{2} + {bc}^{2} + abc + abc + {bc}^{2} + {ac}^{2} = 12$

$\rm :\longmapsto\: 3abc + {ba}^{2} + {ca}^{2}+{ab}^{2} + {bc}^{2}+ {bc}^{2} + {ac}^{2} = 12$

$\rm :\longmapsto\: 3abc + 18 = 12$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (2) \bigg\}}$

$\rm :\longmapsto\: 3abc = 12 - 18$

$\rm :\longmapsto\: 3abc = - 6$

$\rm \implies\:\boxed{\tt{ abc = - 2}} - - - (3)$

Now, Consider

$\rm :\longmapsto\:(ab + bc + ca)^{2} = {3}^{2}$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2( {b}^{2}ac + {2c}^{2}ab + {2bca}^{2}) = 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2abc(a + b + c)= 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2( - 2)(4)= 9$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (3) \: and \: given \bigg\}}$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} - 16= 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} = 9 + 16$

$\rm :\longmapsto\: \boxed{\tt{ {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} = 25 \: }} - - - (4)$

Now, given that

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 10$

On squaring both sides, we get

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 2( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2}) = 100$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 2( 25) = 100$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (4) \bigg\}}$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 50 = 100$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} = 100 - 50$

$\rm \implies\:\boxed{\tt{ \: {a}^{4} +{b}^{4} + {c}^{4} = 50 \: }}$

Answer 2

Answer:

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

Given that

$\rm :\longmapsto\:a + b + c = 4$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 10$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + {c}^{3} = 22$

Now, Consider

$\rm :\longmapsto\:a + b + c = 4$

On squaring both sides, we get

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

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 16$

$\rm :\longmapsto\: 10 + 2(ab + bc + ca) = 16$

$\rm :\longmapsto\: 2(ab + bc + ca) = 16 - 10$

$\rm :\longmapsto\: 2(ab + bc + ca) = 6$

$\rm \implies\:\boxed{\tt{ ab + bc + ca = 3}} - - - (1)$

Now, Consider

$\rm :\longmapsto\:(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2}) = 4 \times 10$

$\rm :\longmapsto\: {a}^{3} + {ab}^{2} + {ac}^{2} +{ba}^{2} +{b}^{3} + {bc}^{2} + {ca}^{2} + {cb}^{2} + {c}^{3} = 40$

$\rm :\longmapsto\: {a}^{3}+ {b}^{3} + {c}^{3} + {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40$

$\rm :\longmapsto\: 22+ {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40$

$\red{ \bigg\{ \sf \: \because \: of \: given \bigg\}}$

$\rm :\longmapsto\:{ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=40 - 22$

$\rm :\longmapsto\:\boxed{\tt{ {ab}^{2}+{ac}^{2} +{ba}^{2} + {bc}^{2} +{ca}^{2} +{cb}^{2}=18}} - - - (2)$

Consider

$\rm :\longmapsto\:(a + b + c)(ab + bc + ca) = 4 \times 3$

$\rm :\longmapsto\: {ba}^{2} +abc + {ca}^{2} + {ab}^{2} + {bc}^{2} + abc + abc + {bc}^{2} + {ac}^{2} = 12$

$\rm :\longmapsto\: 3abc + {ba}^{2} + {ca}^{2}+{ab}^{2} + {bc}^{2}+ {bc}^{2} + {ac}^{2} = 12$

$\rm :\longmapsto\: 3abc + 18 = 12$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (2) \bigg\}}$

$\rm :\longmapsto\: 3abc = 12 - 18$

$\rm :\longmapsto\: 3abc = - 6$

$\rm \implies\:\boxed{\tt{ abc = - 2}} - - - (3)$

Now, Consider

$\rm :\longmapsto\:(ab + bc + ca)^{2} = {3}^{2}$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2( {b}^{2}ac + {2c}^{2}ab + {2bca}^{2}) = 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2abc(a + b + c)= 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} + 2( - 2)(4)= 9$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (3) \: and \: given \bigg\}}$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} - 16= 9$

$\rm :\longmapsto\: {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} = 9 + 16$

$\rm :\longmapsto\: \boxed{\tt{ {a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2} {a}^{2} = 25 \: }} - - - (4)$

Now, given that

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 10$

On squaring both sides, we get

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 2( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2}) = 100$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 2( 25) = 100$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (4) \bigg\}}$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} + 50 = 100$

$\rm :\longmapsto\: {a}^{4} +{b}^{4} + {c}^{4} = 100 - 50$

$\rm \implies\:\boxed{\tt{ \: {a}^{4} +{b}^{4} + {c}^{4} = 50 \: }}$