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If $\frac{4\left[(17)^3-(7)^3\right]}{\left(17^2+7^2+p\right)}=40$, then what is the value of p ? |
-119 -129 119 129 |
119 |
( a - b ) = \(\frac{a^3 - b^3}{a^2 + b^2 + ab }\) $\frac{4\left[(17)^3-(7)^3\right]}{\left(17^2+7^2+p\right)}=40$ By comparing the values from both of the equation we get = a = 17 b = 9 p = ab = 17 × 9 = 119 |
