Question

If a3 + b3 = 604 and a + b = 4, then the value of a

4 + b4

is ​

Answer 1

$\textbf{Given:}$

$a^3+b^3=604\;\text{and}\;a+b=4$

$\textbf{To find:}$

$\text{The value of a^4+b^4 }$

$\textbf{Solution:}$

$\text{Consider,}$

$a^3+b^3=604$

$\implies(a+b)(a^2-ab+b^2)=604$

$\implies(4)(a^2+b^2-ab)=604$

$\implies\,a^2+b^2-ab=151$

$\implies\,(a+b)^2-2ab-ab=151$

$\implies\,(4)^2-3ab=151$

$\implies\,-3ab=151-16$

$\implies\,-3ab=135$

$\implies\,ab=-45$

$\text{Now, we have a+b=4 and ab=-45}$

$\text{By trial and error method, a=9 and b=-5}$

$\bf\,a^4+b^4$

$=9^4+(-5)^4$

$=6561+625$

$=7186$

$\threfore\textbf{The value of \bf\,a^4+b^4 is 7186}$

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Answer 2

To find. $\mathsf{a^{4}+b^{4}}$

Given. $\mathsf{a^{3}+b^{3}=604,\:a+b=4}$

Solution.

Here, $\mathsf{a^{3}+b^{3}=604}$

$\Rightarrow \mathsf{(a+b)^{3}-3ab(a+b)=604}$

$\Rightarrow \mathsf{4^{3}-3ab(4)=604\quad[\because a+b=4]}$

$\Rightarrow \mathsf{64-12ab=604}$

$\Rightarrow \mathsf{12ab=64-604}$

$\Rightarrow \mathsf{12ab=-540}$

$\Rightarrow \underline{\mathsf{ab=-45}}$

Now, $\mathsf{a^{2}+b^{2}}$

$\mathsf{=(a+b)^{2}-2ab}$

$\mathsf{=4^{2}-2(-45)\quad[\because a+b=4,\:ab=-45]}$

$\mathsf{=16+90}$

$\mathsf{=106}$

$\Rightarrow \underline{\mathsf{a^{2}+b^{2}=106}}$

$\therefore \mathsf{a^{4}+b^{4}}$

$\mathsf{=(a^{2}+b^{2})^{2}-2a^{2}b^{2}}$

$\mathsf{=(a^{2}+b^{2})^{2}-2(ab)^{2}}$

$\mathsf{=106^{2}-2(-45)^{2}\quad[\because a^{2}+b^{2}=106,\:ab=-45]}$

$\mathsf{=11236-4050}$

$\mathsf{=7186}$

$\Rightarrow \underline{\mathsf{a^{4}+b^{4}=7186}}$

Answer: $\color{blue}\mathsf{a^{4}+b^{4}=7186}$