Show, with induction that $1^2 + 2^2 + .... + n^2 = frac{n(n+1)(2n+1)}{6}$

Show, with induction that

$1^2 + 2^2 + .... + n^2 = \frac{n(n+1)(2n+1)}{6}$

My attempt

Case 1: n = 1

$LHS = 1^2$

$RHS = \frac{(1+1)(2+1)}{6} = \frac{2*3}{6} = 1$

Case 2: n = p

$LHS_{p} = 1^2 + 2^2 + ... + p^2$

$RHS_{p} = \frac{p(p+1)(2p+1)}{6}$

Case 3: n = p + 1

$LHS_{p+1} = 1^2+2^2+....+p^2+(p+1)^2$

$RHS_{p+1} = \frac{(p+1)((p+1)+1)(2(p+1)+1)}{6}$

Now to show this with induction I think i need to show that

$RHS_{p+1} = RHS_{p} + (p+1)^2$

$RHS_{p+1} = \frac{p(p+1)(2p+1)}{6} + (p+1)^2$

So I need to rewrite

$RHS_{p+1} = \frac{(p+1)((p+1)+1)(2(p+1)+1)}{6}$to be equal to $\frac{p(p+1)(2p+1)}{6} + (p+1)^2$
Anyone see how I can do that? Or got any other solution?