Question

Solve the following equations:
$x+y=1072$,
$x^{\frac{1}{3}}+y^{\frac{1}{3}}=16$.

• A. $x=729;y=343$
• B. $x=720;y=343$
• C. $x=524;y=629$
• D. $x=343;y=729$

Answer 1

Answer: (A), (D)

The correct options are
A $x=729;y=343$
D $x=343;y=729$

$x+y=1072x^{\frac{1}{3}}+y^{\frac{1}{3}}=16$

Put $x=u^3,y=v^3$

$u+v=16$ .......(i)

$u^3+v^3=1072$

Using $a^3+b^3=(a+b)(a^2+b^2-ab)$

$\Rightarrow (u+v)(u^2+v^2-uv)=1072$

Using (i), we get

$16(u^2+v^2-uv)=1072u^2+v^2-uv=67$

Using ${(a+b)}^2=a^2+b^2+2ab$

${(u+v)}^2-3uv=67{\left(16\right)}^2-3uv=67uv=63v=\frac{63}{u}$

From (i), we have

$u+\frac{63}{u}=16\Rightarrow u^2+63=16u\Rightarrow u^2-16u+63=0\Rightarrow u^2-9u-7u+63=0\Rightarrow u(u-9)-7(u-9)=0\Rightarrow (u-7)(u-9)=0\Rightarrow u=7,9$'

Now $v=\frac{63}{u}$

$u=7\Rightarrow v=\frac{63}{7}=9u=9\Rightarrow v=\frac{63}{9}=7$

Now according to our assumption, we have

$x=u^3\Rightarrow x=7^3,9^3\Rightarrow x=343,729$

and $y=v^3$

$\Rightarrow y=9^3,7^3\Rightarrow y=729,343$