Chứng minh
(a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b)(b + c)(c + a)
giải chi tiết giùm nha mình like cho
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Ta có: \(8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)=\frac{1}{6}.48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^8-1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^{16}-1\right)\)
Vì \(7^{16}-1>\frac{1}{6}\left(7^{16}-1\right)\) nên \(7^{16}-1>8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)\)
Ta có: M = xy(x+y) + yz(y+z) + xz (x+z) + 2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(x + y)
= (x + y)(xy + zx + zy + z2)
= (x + y)[x(y + z) + z(y + z)]
M = (x + y)(y + z)(z + x) (đpcm)
Do \(a+b+c=0\Rightarrow a+b=-c\)
Ta có: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(-c\right)^3-3ab\left(-c\right)=\left(-c\right)^3+3abc\)
Do đó: \(a^3+b^3+c^3=\left(-c\right)^3+3abc+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
Kết luận: bạn ghi sai đề.
Ta có: \(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Vì \(\left(a+b+c\right)^3\) \(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)nên \(\left(a+b+c\right)^3-\left(a^3+b^3+c^3\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\Leftrightarrow\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(đpcm\right)\)