Question

Cho a + b + c + d =0. Tính \(a^3+b^3+c^3+d^3-3\left(a+b\right)\left(cd-ab\right)\)

Answer 1

Ta có:

\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)

Do đó: \(\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3-3c^2d-3cd^2-d^2\)

\(\Rightarrow a^3+3ab\left(a+b\right)+b^3=-c^3-3cd\left(c+d\right)-d^2\)

\(\Rightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)

Vì \(a+b=-\left(c+d\right)\) nên

\(\Rightarrow a^3+b^3+c^3+d^3=3cd\left(a+b\right)-3ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=3\left(a+b\right)\left(cd-ab\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3-3\left(a+b\right)\left(cd-ab\right)=0\)

Chúc bạn học tốt!!!

Answer 2

<br class="Apple-interchange-newline"><div id="inner-editor"></div>⇔a+c = -( b+ d)

⇔(a+c)³ = - ( b+d)³

⇔a³ + c³ + 3ac.(a+c) = - [ b³ + d³ + 3bd( b+d) ]

⇔a³ + b³ + c³ + d³ = -3bd(b+d) - 3ac(a+c)

⇔a³+b³+c³+d³= -3bd( b+d) + 3ac( b+d)

⇔a³+b³+c³+d³=3.(ac-bd)(d+b)