Question

If a + b=10 and a2 + b 2 =58 find the value of a3 + b 3​

Answer 1

Answer:

answer is 370 please brainlist me

Answer 2

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

$\rm :\longmapsto\:a + b = 10$

and

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = 58$

$\large\underline{\sf{To\:Find - }}$

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

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

Given that

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = 58$

and

$\rm :\longmapsto\:a + b = 10$

On squaring both sides, we get

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

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + 2ab = 100$

$\rm :\longmapsto\:58 + 2ab = 100$

$\rm :\longmapsto\:2ab = 100 - 58$

$\rm :\longmapsto\:2ab = 42$

$\red{\rm :\longmapsto\:ab = 21}$

Now, again

$\rm :\longmapsto\:a + b = 10$

On cubing both sides, we get

$\rm :\longmapsto\: {(a + b)}^{3} = {10}^{3}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 3ab(a + b) = 1000$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 3 \times 21 \times 10 = 1000$

$\red{\bigg \{ \because \:a + b = 10 \: \: and \: \: \: ab = 21 \bigg \}}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 630 = 1000$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} = 1000 - 630$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} = 370$

Additional Information :-

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

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

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

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

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

$\boxed{ \bf{ \: {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2})}}$

$\boxed{ \bf{ \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2})}}$

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