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\(a^3+b^3+c^3-3abc=\left(a^3+b^3+c^3\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-\left(ab+ac+bc\right)\right)+3abc-3abc\)
\(=\left(a+b+c\right)\left[a^2+b^2+c^2-\left(ab+ac+bc\right)\right]\)
a3 + b3 + c3 - 3abc
= \(\left(a^3+b^3\right)+c^3-3abc\)
= \(\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]+c^3-3abc\) (Đặt x = a + b)
= \(x^3-3abx+c^3-3abc\)
= \(x^3+c^3-3abx-3abc\)
= \(\left(x+c\right)\left(x^2+c^2-cx\right)-3ab\left(x+c\right)\)
= \(\left(x+c\right)\left(x^2+c^2-cx-3ab\right)\) (Thay x = a + b)
= \(\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-c\left(a+b\right)-3ab\right]\)
= \(\left(a+b+c\right)\left(a^2+b^2+2ab+c^2-ca-bc-3ab\right)\)
= \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
a) x4 - y4
= ( x2 - y2 ) ( x2 + y2 )
= ( x - y ) ( x + y ( x2 + y2 )
b) ( a - b ) 3 - ( a - b ) 3
= ( a - b ) 2 ( a - b - a + b )
c) ( a2 + 2ab + b2 ) + ( a + b )3
= ( a + b )2 + ( a +b ) 3
= ( a + b ) 2 ( a + b + 1 )
\(a^3-b^3-c^3-3abc\)
\(=\left(a-b\right)^3-c^3-3abc+3a^2b-3ab^2\)
\(=\left[\left(a-b\right)^3-c^3\right]-3ab\left(c-a+b\right)\)
\(=-\left[c^3-\left(a-b\right)^3\right]-3ab\left(c-a+b\right)\)
\(=-\left(c-a+b\right)\left[c^2+c\left(a-b\right)+\left(a-b\right)^2\right]-3ab\left(c-a+b\right)\)
\(=-\left(c-a+b\right)\left(c^2+ac-cb+a^2-2ab+b^2+3ab\right)\)
\(=-\left(c-a+b\right)\left(c^2+ac-cb+a^2+ab+b^2\right)\)
\(2x^2-9x+8\)
\(=2\left(x^2-\frac{9}{2}x+4\right)\)
Cậu b và c cx z thôi