Given : $$a^2+b^2=32$$ ------------(i)
and $$a+b=8$$ ---------(ii)
Squaring both sides,
=> $$a^2+b^2+2ab=64$$
Substituting value from equation (i), => $$32+2ab=64$$
=> $$2ab=64-32=32$$
=> $$ab=\frac{32}{2}=16$$ ------------(iii)
$$\therefore$$Â $$a^{3}+b^{3}$$
= $$(a+b)(a^2+b^2-ab)$$
Substituting values from equation (i), (ii) and (iii),
= $$(8)(32-16)$$
= $$8\times16=128$$
=> Ans - (C)
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