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