If a-b=3 and a^{3}-b^{3}=117 then find the absolute value of

If a-b=3 and ${a}^{3}-{b}^{3}=117$ then find the absolute value of a+b.
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Step 1
Given that a-b=3 and ${a}^{3}-{b}^{3}=117$.
It is known that ${\left(a-b\right)}^{3}={a}^{3}-{b}^{3}-3ab\left(a-b\right)$.
Given that a-b=3 and ${a}^{3}-{b}^{3}=117$.
${\left(a-b\right)}^{3}={a}^{3}-{b}^{3}-3ab\left(a-b\right)$
${3}^{3}=117-3ab\left(3\right)$
27=117-9ab
9ab=90
ab=10
Step 2
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$
$117=3\left[{\left(a+b\right)}^{2}+ab\right]$
$39={\left(a+b\right)}^{2}+10$
a+b=7