( 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 \(=\left(a+x\right)^3+\left(a-x\right)^3-6ax^2=a^3+3a^2x-3ax^2+x^3+a^3-3a^2x+3ax^2-x^3-6ax^2=2a^3\)

\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2.6=13\)

^{a^2+b^2 = a^2+2ab+b^2-2ab}

              ^{= (a+b)^2 - 2ab}

              ^{= 5^2-2.6= 13}

giả thế nào ấy

chịu thôi chị ơi toán 8 thì sao đây

X^2n - 4 X^n.Y^n-1 + 4Y^2(n-1)

(X ^ n)^2 - 2. X^n.2. Y^n-1 + (2Y ^n-1)^2

= ( X ^N - 2Y^n-1  ) ^2  

(x+y)²=x²+y²+2xy

=>x²+y²=(x+y)²-2xy=5²-2.5

=25-10

=15

(x+y)³=x³+y³+3x²y+3xy²

(x+y)³=x³+y³+3xy(x+y)

=>x³+y³=(x+y)³-3xy(x+y)

=5³-3.5.5

=125-75=50

=>(x²+y²)(x³+y³)=x⁵+x²y³+x³y²+y⁵

(x²+y²)(x³+y³)=x⁵+y⁵+xy(x+y)

=>x⁵+y⁵=(x²+y²)(x³+y³)-xy(x+y)

=15.50-5.5=725

\(\frac{AD}{AB}=\frac{AE}{AC}\)=> DE//BC (TA LÉT)

TAM GIÁC ABI: DH//IB => \(\frac{DH}{IB}=\frac{AH}{AI}\); TAM GIÁC AIC: \(\frac{HE}{IC}=\frac{AH}{AI}\)

=> \(\frac{DH}{IB}=\frac{HE}{IC}\) MÀ IB=IC => HE=HD

(2+1)(2^2+1)(2^4 +1)(2^8+1)(2^16+1) - 2^32

=1.(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2-1).(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2²-1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁴-1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁸-1)(2⁸+1)(2¹⁶+1) - 2³²

=(2¹⁶-1)(2¹⁶+1) - 2³²

=2³²-1-2³²

=-1

    (2+1 ) ( 2^2 + 1) ... (2^16 + 1) - 2^32

=  3 ( 2^2 + 1) ....( 2^16 + 1) -2^32

= ( 2^2 - 1)( 2^2 +1)....(2^16  + 1 ) - 2^32

= (2^4 - 1)( 2^4 + 1)( 2^8 + 1)( 2^16 + 1) - 2^32

= ( 2^8 - 1) ( 2^8 + 1) ( 2^16  - 1 ) - 2^32

= ( 2^ 16 - 1) (2^16 + 1) - 2^32

= 2^32 - 1 - 2^32

=-1