Answer: \(\displaystyle a=2,\quad b=3,\quad c=\frac{544}{5}\).

Answer: \(\displaystyle a=2,\quad b=3,\quad c=\frac{544}{5}\).

First expand the terms needed from

\(\left(ax+\frac25\right)^5.\)

The first three terms in descending powers of \(x\) are

\(a^5x^5+5a^4x^4\left(\frac25\right)+10a^3x^3\left(\frac25\right)^2.\)

So these are

\(a^5x^5+2a^4x^4+\frac85a^3x^3.\)

Also,

The coefficient of \(x^5\) in the full product is \(a^5\). Since this is \(32\),

\(a^5=32,\)

so

\(a=2.\)

Now find the coefficient of \(x^4\). It is

\(2a^4-2ba^5.\)

Using \(a=2\), this becomes

\(2(2^4)-2b(2^5)=32-64b.\)

This is equal to \(-160\), so

\(32-64b=-160.\)

Hence

\(b=3.\)

The coefficient of \(x^3\) is made from three contributions:

\(\frac85a^3-4a^4b+a^5b^2.\)

Using \(a=2\) and \(b=3\),

\(c=\frac85(2^3)-4(2^4)(3)+(2^5)(3^2).\)

Therefore

\(c=\frac{64}{5}-192+288.\)

So

\(c=\frac{544}{5}.\)