[最も選択された] (a-b)^3 formula proof 158706-A^3+b^3+c^3-3abc formula proof
Fundamental Theorem of Calculus x a d F xftdtfx dx where f t is a continuous function on a, x b a f xdx Fb Fa, where F(x) is any antiderivative of f(x) Riemann Sums 11 nn ii ii ca c a 111 nnn ii i i iii ab a b 1A^3b^3 Formula Proof A Cube Plus B Cube Solve a3b3=(ab)(a2b2–ab)= (ab)3=a3b33ab(ab)= (ab)3–3ab(ab)=a3b3= a3b3=(ab)3–3ab(ab) Note Take comA^3 b^3= (ab)^33ab (ab) =>a^3 b^3= (ab) (ab)^23ab =>a^3 b^3= (ab) a^22abb^23ab =>a^3 b^3= (ab) a^2abb^2 this is explanation of first formula hope this is helpful Thanks for the A! Problem If A B C 0 Then Prove That B4 C4 2 B2 C2 Jsunil Tutorial Cbse Maths Science A^3+b^3+c^3-3abc formula proof