# Consider the equation log_10 |x^3 +y^3| - log_10 |x^2-xy +y^2| +log_10 |x^3-y^3| - log _10 |x^2+xy+y^2| = log_10(221). If x+y=17, then the sum of all possible values of x is?

Jul 14, 2018

$15$.

#### Explanation:

We have, ${\log}_{10} \left({x}^{3} + {y}^{3}\right) - {\log}_{10} \left({x}^{2} - x y + {y}^{2}\right)$,

$= {\log}_{10} \left\{\frac{{x}^{3} + {y}^{3}}{{x}^{2} - x y + {y}^{2}}\right\}$,

$= {\log}_{10} \left\{\frac{\left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)}{{x}^{2} - x y + {y}^{2}}\right\}$,

$= {\log}_{10} \left(x + y\right)$.

Similarly, ${\log}_{10} \left({x}^{3} - {y}^{3}\right) - {\log}_{10} \left({x}^{2} + x y + {y}^{2}\right)$

$= {\log}_{10} \left(x - y\right)$.

Utilising these, the given eqn. becomes,

${\log}_{10} \left(x + y\right) + {\log}_{10} \left(x - y\right) = {\log}_{10} 221$.

$\therefore {\log}_{10} \left\{\left(x + y\right) \left(x - y\right)\right\} = {\log}_{10} 221$.

$\therefore \left(x + y\right) \left(x - y\right) = 221. \ldots \ldots \ldots . . \left(1\right)$.

But, we are given that, $x + y = 17. \ldots \left(2\right)$.

Then, by $\left(1\right) , x - y = \frac{221}{17} , i . e . , x - y = 13. \ldots \left(3\right)$.

Solving $\left(2\right) \mathmr{and} \left(3\right) , x = 15$.

Thus, the only possible values of $x$ for which the eqn. holds is

$x = 15$.

Hence, the desired sum is $15$.