154. Eliminate $x$ and $y$ from the equations:

$$
a=x+y, \quad b=x^{3}+y^{3}, \quad c=x^{5}+y^{5}
$$
$\triangle$ Let's raise the first equality to the third power and subtract the second equality:

$$
a^{3}-b=(x+y)^{3}-x^{3}-y^{3}=3 x y(x+y)=3 a x y, \quad x y=\frac{a^{3}-b}{3 a}
$$

Let's raise the first equality to the second power and multiply by the second equality:

$a^{2} b=(x+\mathrm{y})^{2}\left(x^{3}+y^{3}\right)=\left(x^{2}+2 x y+y^{2}\right)\left(x^{3}+y^{3}\right)=$

$=x^{5}+x^{2} y^{3}+y^{2} x^{3}+y^{5}+2 x y\left(x^{3}+y^{3}\right)=$

$=\left(x^{5}+y^{5}\right)+x^{2} y^{2}(x+y)+2 x y\left(x^{3}+y^{3}\right)=$ $=c+\left(\frac{a^{3}-b}{3 a}\right)^{2} a+2 \frac{a^{3}-b}{3 a} b$.

It remains to simplify the obtained equality:

$$
a^{2} b=c+\left(\frac{a^{3}-b}{3 a}\right)^{2} a+2 \frac{a^{3}-b}{3 a} b
$$

Answer: $5 b\left(a^{3}+b\right)=a\left(a^{5}+9 c\right)$.