So the Question states : Show that $1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2$
This is what I have tried so far:
Let $n=1$, $1^3=1^2$, $1=1$, which is true
$1^3+2^3+3^3+...+k^3=(1+2+3+...+k)^2$ Is True
So
$1^3+2^3+3^3+...+(k+1)^3=(1+2+3+...+k+1)^2$
$1^3+2^3+3^3+...+k^3+3k^2+3k+1=(1+2+3+...+k+1)^2$
$(1+2+3+...+k)^2+3k^2+3k+1=(1+2+3+...+k+1)^2$
I don't really know what to do next, Sorry for the layout of the question, I couldn't master the format
Assuming that$$1^3+2^3+\cdots+k^3=(1+2+\cdots+k)^2,$$then$$1^3+2^3+\cdots+k^3+(k+1)^3=(1+2+\cdots+k)^2+(k+1)^3.$$Is this equal to $\bigl(1+2+\cdots+(k+1)\bigr)^2$? This is the same thing as asserting that$$(k+1)^2=\bigl(1+2+\cdots+(k+1)\bigr)^2-(1+2+\cdots+k)^2.$$But the right-hand side of this equality is equal to\begin{align}(k+1)\times\bigl(2\times(1+2+\cdots+k)+(k+1)\bigr)&=(k+1)\times\bigl(k(k+1)+(k+1)\bigr)\\&=(k+1)^3.\end{align}
Hypothesis: True for $n.$
$1^3+2^3+....+n^3 = $
$(1+2+3 ..+ n)^2$
Step:
LHS: $1^3+2^3+3^3 ....+ n^3 + (n+1)^3$
RHS: $(\dfrac{n(n+1)}{2})^2+(n+1)^3 =$
$(1/4)((n+1)^2(n^2 +4(n+1))=$
$(1/4)(n+1)^2(n+2)^2=$
$[(1/2)(n+1)(n+2)]^2=$
$(1+2+3...+n+(n+1))^2$.
Used: $1+2+3....n = \dfrac{n(n+1)}{2}.$
