find minimal vvalue of the expression a^6+b^6 if a^2+b^2=1?( if ur answer is 1/4 tell me how 3a^2b^2 becomes 3/4)

We have to minimize $$a^{6}+b^{6}$$
since $$x^{3}+y^{3}=(x+y)(x^{2}+y^{2}-xy)$$
So,
$$a^{6}+b^{6}=(a^{2}+b^{2})(a^{4}+b^{4}-a^{2}b^{2})$$
$$(a^{2}+b^{2})^{2}=1$$
$$(a^{4}+b^{4}-a^{2}b^{2})+3a^{2}b^{2}=1$$
which is $$1-3a^{2}b^{2}$$
to minimize this expression $$a^{2}b^{2}$$ should be maximum.
We know that $$a^{2}+b^{2}=1$$
Let a^2=m and b^2=n
m+n=1
GM<=AM
$$\sqrt{mn}$$<=$$\frac{m+n}{2}$$
This means, $$mn$$<=$$(\frac{1}{2})^{2}$$
this means its maximum value is 1/4.
thus minimum value of expression becomes- 1-3/4=1/4.