Question

Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which ${\log }_{10}(x+y)=z$ and ${\log }_{10}\left(x^2+y^2\right)=z+1$. Suppose there are real numbers $a$ and $b$ such that for all ordered triples $(x,y,z)$ in $S$ we have $x^3+y^3=a\cdot 10^{3z}+b\cdot 10^{2z}$. The value of $(a+b)$ is equal to

• A. $\frac{15}{12}$
• B. $\frac{29}{2}$
• C. 15
• D. 24

Answer 1

Answer: (B)

$x+y=10^z$....(1) and
$x^2+y^2=10\cdot 10^z$...(2) (given)
now
$(x+y{)}^2=10^{2z}\text{ squaring }(1)$
$x^2+y^2+2xy=10^{2z}$
$2xy=10^{2z}-10\cdot 10^z$....(3)
$\text{ now, }{x}^3+y^3=(x+y{)}^3-3xy(x+y)=(10{)}^{3z}-\frac{3}{2}\left\{(10{)}^{2z}-10\cdot (10{)}^z\right\}10^z$
$=(10{)}^{3z}-\frac{3}{2}\cdot (10{)}^{3z}+15\cdot (10{)}^{2z}=-\frac{1}{2}(10{)}^{3z}+15\left(10^{2z}\right)$
hence $a=-\frac{1}{2}$ and $b=15\Rightarrow a+b=\frac{29}{2}$