Question

chứng minh rằng:

a, (a+b)(a²-ab+b²) + (a-b)(a²+ab+b²) = 2a³.

b, a³+b³= (a+b)[(a-b)²+ab]

c, (a²+b²)(c²+d²)= (ac+bd)²+ (ad-bc)²

Answer 1

a, Ta có: \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)

= \(a^3+b^3+a^3-b^3=a^3+a^3=2a^3\)

\(\xrightarrow[]{}\) đpcm

b, Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(\left(a-b\right)^2+ab\right)\)

\(\xrightarrow[]{}\) đpcm

c, Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(\xrightarrow[]{}\) đpcm

Answer 2

Tham khảo nè!!

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Answer 3

\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a+ab+b^2\right)=a^2+b^3+a^3-b^3=2a^3\)\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\Rightarrowđpcm\)\(c,\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+a^2d^2+b^2d^2+b^2c^2=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\Rightarrowđpcm\)