(x-y)³+(y-z)³+(z-x)³ / (x²-y²)³+(y²-z²)³+(z²-x²)³ = ?
Answers
Answer:
(x-y)³+(y-z)³+(z-x)³/(x²-y²)³+(y²-z²)³+(z²-x²)³=?.
(x³-y³-3xy(x-y))+(y³-z³-3yz(y-z))+(z³-x³-3zx(z-x))/((x²)³-(y²)³-3x²y²(x²-y²))+((y²)³-(z²)³-3y²z²(y²-z²))+((z²)³-(x²)³-3z²x²(z²-x²)).
(Because (a-b)³=a³-b³-3ab(a-b)).
(x³-y³-3x²y+3xy²)+(y³-z³-3y²z+3yz²)+(z³-x³-3z²x+3zx²)/(x^6-y^6-3x⁴y²+3x²y⁴)+(y^6-z^6-3y⁴z²+3y²z⁴)+(z^6-x^6-3z⁴x²+3z²x⁴).
(Because -×-=+ & (a^m)^n=a^m×n).
x³-y³-3x²y+3xy²+y³-z³-3y²z+3yz²+z³-x³-3z²x+3zx²/x^6-y^6-3x⁴y²+3x²y⁴+y^6-z^6-3y⁴z²+3y²z⁴+z^6-x^6-3z⁴x²+3z²x⁴.
(Because +×-=-).
3xy²-3x²y+3yz²-3y²z+3zx²-3z²x/3x²y⁴-3x⁴y²+3y²z⁴-3y⁴z²+3z²x⁴-3z⁴x².
(Because +×-=- & -×+=- so, x³-x³, -y³+y³, -z³+z³, x^6-x^6, -y^6+y^6 & -z^6+z^6 will cancel i.e., it will be zero(0)).
1/xy²+1/x²y+1/yz²+1/y²z+1/zx²+1/z²x.
(Because 3xy²/3x²y⁴=1/xy², -3x²y/-3x⁴y²=1/x²y, 3yz²/3y²y⁴=1/yz², -3y²z/-3y⁴z²=1/y²z, 3zx²/3z²x⁴=1/zx² & -3z²x/-3z⁴x²=1/z²x & -÷-=+).
1(x²y)/1(xy²)+1(y²z)/1(yz²)+1(z²x)/1(zx²).
(Because by cross multiplying).
x²y/xy²+y²z/yz²+z²x/zx².
I think this is your answer.