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a ) \(\left(x+y\right)^3+\left(x-y\right)^3-2x^3\)
\(=x^3+3x^2y+3y^2x+y^3+x^3-3x^2y+3y^2x-y^3-2x^3\)
\(=\left(x^3+x^3-2x^3\right)+\left(y^3-y^3\right)+\left(3x^2y-3x^2y\right)+\left(3y^2x+3y^2x\right)\)
\(=6y^2x\)
b ) \(\left(x+y\right)^2-\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)+x^2-y^2\)
\(=2y.2x+x^2-y^2\)
\(=x^2-y^2+4xy\)
c ) \(\left(3x+1\right)^2+2\left(9x^2-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1\right)^2+2\left(3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1+3x-1\right)^2\)
\(=\left(6x\right)^2=36x^2\)
d ) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2\)
\(=a^2\)
a) \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)
\(=\left(x^3-2x^2-2x^2+4x+2x-4\right)\left(x^3+2^3\right)\)
\(=\left(x^3-4x^2+6x-4\right)\left(x^3+8\right)\)
\(=x^6+8x^3-4x^5-32x^2+6x^4+48x-4x^3-32\)
\(=x^6-4x^5+4x^3-32x^2+48x-32\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]+x^3-3x\left(x^2-1\right)\)
\(=2x\left[\left(x^2+2x+1\right)-\left(x^2-1\right)+\left(x^2-2x+1\right)\right]+x^3-\left(3x^3-3x\right)\)
\(=2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)+x^3-3x^3+3x\)
\(=2x\left(x^2+3\right)+x^3-3x^3+3x\)
\(=2x^3+6x-2x^3+3x\)
\(=9x\)
2 câu kia đợi tí đã nhé!
c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)
\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(4a^2-4ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)
\(=6a^2+3b^2+2c^2\)
d) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+2a^2+2ab+b^2\)
\(=4a^2+4b^2+2c^2+6ab.\)
1.
Đặt $x^2+y^2=a; z^2-x^2=b$ thì $y^2+z^2=a+b$
$(x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3=a^3+b^3-(a+b)^3$
$=a^3+b^3-[a^3+b^3+3ab(a+b)]$
$=-3ab(a+b)=-3(x^2+y^2)(z^2-x^2)(y^2+z^2)$
$=3(x^2+y^2)(x-z)(x+z)(y^2+z^2)$
2.
$a(b+c)^2(b-c)+b(c+a)^2(c-a)+c(a+b)^2(a-b)$
$=(ab+ac)(b^2-c^2)+(bc+ba)(c^2-a^2)+(ca+cb)(a^2-b^2)$
$=(ab+ac)(b^2-c^2)-(bc+ba)[(b^2-c^2)+(a^2-b^2)]+(ca+cb)(a^2-b^2)$
$=(b^2-c^2)(ab+ac-bc-ba)+(a^2-b^2)(ca+cb-bc-ba)$
$=(b^2-c^2)(ac-bc)+(a^2-b^2)(ca-ba)$
$=(b-c)(b+c)c(a-b)-(a-b)(a+b)a(b-c)$
$=(a-b)(b-c)[c(b+c)-a(a+b)]$
$=(a-b)(b-c)[b(c-a)+(c^2-a^2)]=(a-b)(b-c)(c-a)(b+c+a)$
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Đặt \(a+b-c=x , b+c-a=y , c+a-b=z\)
Suy ra biểu thức \(=\left(x+y+z\right)^3-x^3-y^3-z^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Thế vô là xong