( a + b + c)³ +  (a - b - c)³ - 6a( b+ c)²

= ( a+ b + c + a - b - c)[ (a+b+c)² + (a+b+c)(a-b-c) + (a-b-c)² ] - 6a( b+c)²

= 2a [ a² + b² + c² + 2ab+ 2bc+ 2ac + a² - ( b+ c)² + a² + b² + c² - 2ab - 2ac + 2bc] - 6a ( b+c)²

= 2a [ 3a² + 2b² + 2c² + 4bc - (b+c)² - 3(b+c)²}

= 2a ( 3a² + 2b² + 2c² - 4( b+ c)² + 4bc} 

Đặt \(b+c=x\)

Biểu thức đã cho \(=\left(a+x\right)^3+\left(a-x\right)^3-6ax^2=a^3+3a^2x-3ax^2+x^3+a^3-3a^2x+3ax^2-x^3-6ax^2=2a^3\)

ko bits

\(A=\left(x+y+z\right)^3-\left(x+y-z\right)^3-\left(y+z-x\right)^3+\left(z+x-y\right)^3\)

Đặt \(B=\left(x+y+z\right)^3-\left(x+y-z\right)^3\)

\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)\cdot z^2+z^3-\left(x+y\right)^3+3z\left(x+y\right)^2-3\left(x+y\right)\cdot z^2+z^3\)

\(=6z\left(x+y\right)^2+2z^3\)

\(C=-\left(y+z-x\right)^3+\left(z+x-y\right)^3\)

\(=\left(x-y+z\right)^3+\left(x-y-z\right)^3\)

\(=\left(x-y\right)^3+3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2+z^3+\left(x-y\right)^3-3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2-z^3\)

\(=2\left(x-y\right)^3+6\left(x-y\right)\cdot z^2\)

=>\(A=6z\left(x+y\right)^2+2z^3+2\left(x-y\right)^3+6z^2\left(x-y\right)\)

a) (-a+b)-(-a+c)=-a+b+a-c=b-c

-a+b-a+c

-a+a-b+c

0-b+c

b+c

\(A=\log_a\left(a^2\sqrt[4]{a^3\sqrt[5]{a}}\right)=\log_a\left(a^2\sqrt[4]{a^3.a^{\frac{1}{5}}}\right)=\log_a\left[a^2\left(a^{\frac{16}{5}}\right)^{\frac{1}{4}}\right]=\log_a\left(a^2.a^{\frac{4}{5}}\right)=\frac{14}{5}\)

\(M=lg\left|\log_{\frac{1}{a^3}}\sqrt[5]{a\sqrt{a}}\right|=lg\left|\log_{\frac{1}{a^3}}\sqrt[5]{a.a^{\frac{1}{2}}}\right|=lg\left|\log_{\frac{1}{a^3}}\left(a^{\frac{3}{2}}\right)^{\frac{1}{5}}\right|=lg\left|\log_{a^{-3}}a^{\frac{3}{10}}\right|=lg\left|-\frac{1}{10}=lg\frac{1}{10}=-1\right|\)

đ/s:180875.8725