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Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
1)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(b2c-b2a)+(c2a-a2c)
=b.(a2-c2)-b2.(a-c)-ac.(a-c)
=b.(a-c)(a+c)-b2(a-c)-ac(a-c)
=(a-c)(ab+bc-b2-ac)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
\(\left(a-b+c\right)^2+\left(c-b\right)^2+2\left(a-b+c\right)\left(b-c\right)\)
\(=\left(a-b+c\right)\left[a-b+c+2\left(b-c\right)\right]+\left(c-b\right)^2\)
\(=\left(a-b+c\right)\left[a-b+c+2b-2c\right]+\left(c-b\right)^2\)
\(=\left(a-b+c\right)\left[a+b-c\right]+\left(c^2-2bc+b^2\right)\)
\(=-c^2+2bc-b^2+a^2\)\(+\left(c^2-2bc+b^2\right)\)
\(=a^2\)
\(\left(a^2+a+4\right)^2+8a\left(a^2+a+4\right)+15a^2=\left(a^2+a+4\right)^2+8a\left(a^2+a+4\right)+16a^2-a^2=\left(a^2+a+4+4a\right)^2-a^2\)
\(\left(a^2+5a+4\right)^2-a^2=\left(a^2+5a+a+4\right)\left(a^2+5a-a+4\right)=\left(a^2+6a+4\right)\left(a^2+4a+4\right)=\left(a^2+6a+4\right)\left(a+2\right)^2\)
a,9a^3-13a+6
=9a^3-6a^2+6a^2-4a-9a+6
=3a^2(3a-2)+2a(3a-2)-3(3a-2)
=(3a^2+2a-2)(3a-2)