It is true $\dots$ or is it?

$\begin{aligned} \left(\frac {a(a+1)}2\right)^2 - \left(\frac {a(a-1)}2\right)^2 & = \frac {a^2}4 ((a+1)^2 - (a-1)^2) \\ & = \frac {a^2}4 \left(a^2+2a+1 - (a^2 - 2a +1)\right) \\ & = \frac {a^2}4(4a) \\ & = a^3 \quad \boxed{\text{True}} \end{aligned}$

$a^3=(\frac{a(a+1)}{2})^2-(\frac{a(a-1)}{2})^2$

Because $(\frac{a(a+1)}{2})^2-(\frac{a(a-1)}{2})^2=\frac{a^2(a+1)^2-a^2(a-1)^2}{4}=\frac{a^2 \times 4a}{4}=a^3$ so the answer is True

$\begin{aligned} \left\{\dfrac {a(a+1)}2\right\}^2- \left\{\dfrac {a(a-1)}2 \right\}^2 & = \left(\dfrac{a^2+a}2\right)^2-\left(\dfrac{a^2-a}2\right)^2 \\ & = a^2 \cdot a \\ & = a^3 \\ \end{aligned}$

Note: $ab = \left(\dfrac{a+b}2 \right)^2 - \left(\dfrac{a-b}2 \right)^2$