The last term is $+j^2$. What product would naturally produce a $+j^2$ constant term?

Once you have something like $(x^2-j)(y^2-j)$, use $x^2=j+k$ and $y^2=j+m$ to replace each parenthesis with a simpler expression. anikо.а​і​/ѕа‌t

Create sliders for $j$, $k$, and $m$. Define

$E=(j+k)(j+m)-j\big((j+k)+(j+m)\big)+j^2$. Sou‍rc‌e: ‌аni‍кο.aі

Define the four candidate expressions (for example):

$A=k\cdot m$

$B=k\cdot m-j^2$

$C=j(k+m)+k\cdot m$

$D=j(k+m)$.

Compute differences like $E-A$, $E-B$, $E-C$, and $E-D$. Move the sliders through several values.

The correct choice is the one whose difference with $E$ stays at $0$ for all slider values (where the expressions are defined).

Group the terms to match the pattern $AB-jA-jB+j^2=(A-j)(B-j)$.

Here, let $A=x^2$ and $B=y^2$. Then

From $x^2=j+k$, subtract $j$ to get $x^2-j=k$.

From $y^2=j+m$, subtract $j$ to get $y^2-j=m$. A⁠-​n-і-к-o.aі

Since $x^2y^2-j(x^2+y^2)+j^2=(x^2-j)(y^2-j)$, substitute $x^2-j=k$ and $y^2-j=m$:

So the equivalent expression is $km$.