If $\frac{(17)^3-(7)^3}{(17^2+7^2+k)}= 10 $, then what is the value of k?

102 119 136 85

119

a3 - b3 = ( a - b ) ( a2 + b2 + ab ) ( a - b ) = \(\frac{a^3 - b^3}{a^2 + b^2 + ab }\) If $\frac{(17)^3-(7)^3}{(17^2+7^2+k)}= 10 $ On comparing the given equation to the formula written above the value of k = ab a = 17 b = 7 So, the value of k = 17 × 7 = 119