If a-b=3 and a3−b3=117 then find the absolute value of a+b.

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Joseph Fair · Accepted Answer

Step 1  Given that a-b=3 and a3−b3=117.  It is known that (a−b)3=a3−b3−3ab(a−b).  Given that a-b=3 and a3−b3=117.  (a−b)3=a3−b3−3ab(a−b)  33=117−3ab(3)  27=117-9ab  9ab=90  ab=10  Step 2  a3−b3=(a−b)(a2+b2+ab)  117=3[(a+b)2+ab]  39=(a+b)2+10  a+b=7

If a-b=3 and a3−b3=117 then find the absolute value of a+b.

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