\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+b\left(c^3-b^3+b^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+b\left(c^3-b^3\right)+b\left(b^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)

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

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

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

\(=\left(b-c\right)\left(a-b\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)

vì ngọn lửa biểu tượng cho niềm tin ,khát vọng

==>ở đây'nhà văn muốn nói lên những khát khao của cô bé

\(\hept{\begin{cases}f\left(x\right)=x^4+6x^2+25⋮P\left(x\right)\\g\left(x\right)=3x^4+4x^2+28x+5⋮P\left(x\right)\end{cases}}\)

\(\Rightarrow3f\left(x\right)-g\left(x\right)=3\left(x^4+6x^2+25\right)-\left(3x^4+4x^2+28x+5\right)=14\left(x^2-2x+5\right)⋮P\left(x\right)\)

\(\Rightarrow P\left(x\right)=x^2-2x+5\)

Thử lại: \(f\left(x\right)=\left(x^2+2x+5\right)\left(x^2-2x+5\right)⋮P\left(x\right)\)

              \(g\left(x\right)=\left(3x^2+6x+1\right)\left(x^2-2x+5\right)⋮P\left(x\right)\)

do đó \(P\left(x\right)=x^2-2x+5\)thỏa mãn. 

\(P\left(-2\right)=\left(-2\right)^2-2.\left(-2\right)+5=13\)

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

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\).

\(A=a^3+b^3+c^3-3abc+3ab-3c+5\)

\(A=a^3+a^3+a^3-3a^3+3a^2-3a+5\)

\(A=3a^2-3a+5\)

\(A=3\left(a-\frac{1}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\).

Dấu \(=\)khi \(a=\frac{1}{2}\).

Vậy GTNN của \(A\)là \(\frac{17}{4}\).

Đặt \(\hept{\begin{cases}a=2^x-8\\b=4^x+13\end{cases}}\).

Ta có: \(a^3+b^3=\left(a+b\right)^3\)

 \(\Leftrightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Leftrightarrow ab\left(a+b\right)=0\)

Ta có các trường hợp:

- \(a=0\Rightarrow2^x-8=0\Leftrightarrow2^x=2^3\Leftrightarrow x=3\).

- \(b=0\Rightarrow4^x+13=0\)(vô nghiệm)

- \(a+b=0\Rightarrow4^x+2^x+5=0\)(vô nghiệm) 

 (do \(4^x,2^x>0\)với mọi \(x\inℝ\))

a, \(x^2\left(x-y\right)+y^2\left(y-x\right)=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\)

b, \(x^2+5x+6=x^2+3x+2x+6=\left(x+2\right)\left(x+3\right)\)