Toán 8 nâng cao:

M

motngaymoi12

H

hotien217

1

$(x+y+z)^3 + (x-y-z)^3 $
$= x^3+3 x^2 y+3 x^2 z+3 x y^2+6 x y z+3 x z^2+y^3+3 y^2 z+3 y z^2+z^3$
$+ x^3-3 x^2 y-3 x^2 z+3 x y^2+6 x y z+3 x z^2-y^3-3 y^2 z-3 y z^2-z^3$
$= 2 x^3+6 x y^2+12 x y z+6 x z^2$
$= 2 x^3+6 x (y+z)^2$
P/s: Mình chỉ khai triển rồi rút gọn chứ không có cách làm hay. Bạn nào có cho ình tham khảo :)
 
I

iceghost

Đặt $x+y+z=a, x-y-z=b$
Ta có : $a+b=x+y+z+x-y-z=2x$
$\implies$ $3(a+b)=6x \\
\dfrac{(a+b)^3}{8}=x^3 \iff \dfrac{(a+b)(a+b)^2}{4}=2x^3$

Lại có : $a-b=x+y+z-(x-y-z)=2y+2z=2(y+z)$
$\implies$ $\dfrac{(a-b)^2}4=(y+z)^2$

Mặt khác : $B=6x(y+z)^2+2x^3 \\
\iff 3(a+b).\dfrac{(a-b)^2}4+\dfrac{(a+b)(a+b)^2}{4} \\
=\dfrac{3(a+b)(a-b)^2+(a+b)(a+b)^2}4 \\
=\dfrac{(a+b)[3(a-b)^2+(a+b)^2]}4 \\
=\dfrac{(a+b)(3a^2-6ab+3b^2+a^2+2ab+b^2)}4 \\
=\dfrac{(a+b)(4a^2-4ab+4b^2)}4 \\
=\dfrac{4(a+b)(a^2-ab+b^2)}4 \\
=(a+b)(a^2-ab+b^2) = a^3 + b^3 = (x+y+z)^3+(x-y-z)^3 = A$
 
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