\(a,A=\left(x+5\right)^3=\left(-10+5\right)^3=\left(-5\right)^3=-125\\ b,B=\left(2x+3y\right)^2=\left(2\cdot1+3\cdot2\right)^2=7^2=49\\ c,C=\left(3x-y\right)^3=\left(3\cdot1+2\right)^3=5^3=125\)

\(a,\Leftrightarrow\left(x^2-3\right)^2=0\\ \Leftrightarrow x^2-3=0\\ \Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\\ b,\Leftrightarrow8x^3+12x^2+6x+1-64=0\\ \Leftrightarrow\left(2x+1\right)^3-4^3=0\\ \Leftrightarrow\left(2x+1-4\right)\left[\left(2x+1\right)^2+4\left(2x+1\right)+16\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=3\\4x^2+4x+1+8x+4+16=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\4x^2+12x+17=0\left(1\right)\end{matrix}\right.\)

Xét \(\left(1\right)\Leftrightarrow\left(2x+3\right)^2+8=0\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

Vậy pt có nghiệm \(x=\dfrac{3}{2}\)

\(c,\Leftrightarrow\left(3-2x-5\right)\left(3-2x+5\right)=0\\ \Leftrightarrow\left(-2-2x\right)\left(8-2x\right)=0\\ \Leftrightarrow-2\left(x+1\right)\cdot2\left(4-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\) 

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

Pt $\Leftrightarrow (x-4)^3=0\\\Leftrightarrow x-4=0\\\Leftrightarrow x=4$

⇔x³-3.x².4+3.x.4²-4³=0

⇔(x-4)³=0

⇔x-4=0

⇔x=4

\(a,=\left(3x+2y\right)^3\\ b,=\left(4-x\right)^3\\ c,=\left(\dfrac{1}{2}x-3y\right)^3\)