Sum of products of multiplicative characters of $\mathbb{F}_q$ This is problem 5.30 from the book "Introduction to Finite Fields" by Lidl and Niederreiter.

Let $\lambda_1, \lambda_2, \lambda_3$ be nontrivial multiplicative characters of $\mathbb{F}_q$  and let $a_1\not=a_2\in\mathbb{F}_q$. Prove that
$$
\sum_{b\in\mathbb{F}_q} \:\left | \: \sum_{c\in\mathbb{F}_q} \lambda_1(c+a_1)\lambda_2(c+a_2)\lambda_3(c+b) \:\right| ^2  = \begin{cases}{q^2 - 3q  \:\:\:\:\:\:\:\:\:\:\text{ if $\lambda_1\lambda_2$ nontrivial,} \\
q^2 - 2q - 1 \:\:\:\text{ if $\lambda_1\lambda_2$ trivial.} }
\end{cases}
$$

I can reduce the above formula to
$$
q^2-2q - \sum_{c\in\mathbb{F}_q} \sum_{d\in\mathbb{F}_q} \lambda_1(c+a_1)\lambda_2(c+a_2)\overline{\lambda_1(d+a_1)\lambda_2(d+a_2)}.
$$
If $\lambda_1\lambda_2$ is trivial, then the sum is equal to $(-1)^2$ so we obtain the wanted result.
However if $\lambda_1\lambda_2$ is nontrivial, we have , we can simplify the double sum above to
$$
\left | \: \sum_{c\in\mathbb{F}_q} \lambda_1(c+a_1)\overline{\lambda_2^{-1}(c+a_2)} \: \right | ^2_.
$$
This is awfully similar to the product of two multiplicative characters which states
$$
\left | \:\sum_{c\in\mathbb{F}_q} \lambda(c)\overline{\psi(c)} \: \right | ^2 = q-1 \quad \text{ if $\lambda \psi^{-1}$ nontirvial.}
$$
This would indeed lead to the wanted result where $\lambda=\lambda_1$ and $\psi=\lambda_2^{-1}$, however there is an additive factor of $a_1$ (resp. $a_2$) in the argument of $\lambda_1$ (resp. $\lambda_2^{-1}$) which doesn't allow me to use the above formula.
 A: I couldn't find discussion of them in the book you're using, but Jacobi sums are close cousins of Gauss sums that solve this problem, namely they have similar absolute value properties, and deal with additivity. One can read a discussion of them (over a field with $q$ elements) here https://www.math.mcgill.ca/goren/SeminarOnCohomology/mycohomologytalk.pdf
Either way, the result I'll be needing to finish simplifying your sum is that for $\lambda_1\neq \lambda_2^{-1}$ multiplicative characters of $\mathbb{F}_q$, the following expression has absolute value $|q|^{1/2}:$\begin{equation}J(\lambda_1,\lambda_2):=\sum_{c\in\mathbb{F}_q}\lambda_1(c)\lambda_2(1-c)\end{equation}
So it remains to show that your simplification of that double sum can be simplified to a Jacobi sum of this form. Your sum is:
$$\: \sum_{c\in\mathbb{F}_q} \lambda_1(c+a_1)\overline{\lambda_2^{-1}(c+a_2)} \:$$
We have that $\lambda_2^{-1}(c)=\overline{\lambda_2(c)}$, so your sum is:
$$\: \sum_{c\in\mathbb{F}_q} \lambda_1(c+a_1)\lambda_2(c+a_2) \:$$
Re-indexing our summation by the substitution $d:=c+a_1$ yields:
$$\: \sum_{d\in\mathbb{F}_q} \lambda_1(d)\lambda_2(d-a_1+a_2) \:$$
Since $a_1\neq a_2$, we can pull out common factors to obtain:
$$\: \lambda_1(a_1-a_2)\lambda_2(a_2-a_1)\sum_{d\in\mathbb{F}_q} \lambda_1(\frac{d}{a_1-a_2})\lambda_2(1-\frac{d}{a_1-a_2}) \:$$
So then multiplicatively re-indexing our sum by $e:=\frac{d}{a_1-a_2}$ we reach the desired
$$\: \lambda_1(a_1-a_2)\lambda_2(a_2-a_1)\sum_{e\in\mathbb{F}_q} \lambda_1(e)\lambda_2(1-e) \:=\lambda_1(a_1-a_2)\lambda_2(a_2-a_1)J(\lambda_1,\lambda_2)$$
From which taking absolute values gives the result.
