Exercise 5.2 — Question 2(e)
Simplify:
\[
(x+y+z)^2+
\left(x+\frac{y}{2}+\frac{z}{3}\right)^2-
\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^2
\]
Solution
We use the identity:
\[
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca
\]
Step 1: Expand the first square
\[
(x+y+z)^2
=x^2+y^2+z^2+2xy+2yz+2xz
\]
Step 2: Expand the second square
\[
\begin{aligned}
\left(x+\frac{y}{2}+\frac{z}{3}\right)^2
&=
x^2+\frac{y^2}{4}+\frac{z^2}{9}
+2\left(x\right)\left(\frac{y}{2}\right)\\
&\quad
+2\left(\frac{y}{2}\right)\left(\frac{z}{3}\right)
+2\left(x\right)\left(\frac{z}{3}\right)\\
&=x^2+\frac{y^2}{4}+\frac{z^2}{9}
+xy+\frac{yz}{3}+\frac{2xz}{3}
\end{aligned}
\]
Step 3: Expand the third square
\[
\begin{aligned}
\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^2
&=
\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}\\
&\quad
+2\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)
+2\left(\frac{y}{3}\right)\left(\frac{z}{4}\right)
+2\left(\frac{x}{2}\right)\left(\frac{z}{4}\right)\\
&=
\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}
+\frac{xy}{3}+\frac{yz}{6}+\frac{xz}{4}
\end{aligned}
\]
Step 4: Substitute the expansions
\[
\begin{aligned}
&x^2+y^2+z^2+2xy+2yz+2xz\\
&\quad+x^2+\frac{y^2}{4}+\frac{z^2}{9}
+xy+\frac{yz}{3}+\frac{2xz}{3}\\
&\quad-\frac{x^2}{4}-\frac{y^2}{9}-\frac{z^2}{16}
-\frac{xy}{3}-\frac{yz}{6}-\frac{xz}{4}
\end{aligned}
\]
Step 5: Collect like terms
For \(x^2\):
\[
x^2+x^2-\frac{x^2}{4}
=\frac{7x^2}{4}
\]
For \(xy\):
\[
2xy+xy-\frac{xy}{3}
=\frac{8xy}{3}
\]
For \(xz\):
\[
2xz+\frac{2xz}{3}-\frac{xz}{4}
=\frac{29xz}{12}
\]
For \(y^2\):
\[
y^2+\frac{y^2}{4}-\frac{y^2}{9}
=\frac{13y^2}{36}
\]
For \(yz\):
\[
2yz+\frac{yz}{3}-\frac{yz}{6}
=\frac{13yz}{6}
\]
For \(z^2\):
\[
z^2+\frac{z^2}{9}-\frac{z^2}{16}
=\frac{151z^2}{144}
\]
Final Answer
\[
\boxed{
\frac{7x^2}{4}
+\frac{8xy}{3}
+\frac{29xz}{12}
+\frac{13y^2}{36}
+\frac{13yz}{6}
+\frac{151z^2}{144}
}
\]