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人気のある代数 >

solvefor x,(x+y+z)^3=x^3+y^3+z^3

解

解く

x, (x + y + z)^{3} = x^{3} + y^{3} + z^{3}

解

x = - z, x = - y; 3 y + 3 z \neq = 0

解答ステップ

(x + y + z)^{3} = x^{3} + y^{3} + z^{3}

拡張

(x + y + z)^{3} : y^{3} + 3 x y^{2} + 3 y^{2} z + 3 x^{2} y + 3 y z^{2} + 6 x yz + x^{3} + z^{3} + 3 x z^{2} + 3 x^{2} z

y^{3} + 3 x y^{2} + 3 y^{2} z + 3 x^{2} y + 3 y z^{2} + 6 x yz + x^{3} + z^{3} + 3 x z^{2} + 3 x^{2} z = x^{3} + y^{3} + z^{3}

z^{3}

を左側に移動します

3 x z^{2} + 3 y z^{2} + 3 x^{2} z + 3 y^{2} z + 6 x yz + x^{3} + y^{3} + 3 x y^{2} + 3 x^{2} y = x^{3} + y^{3}

y^{3}

を左側に移動します

3 x y^{2} + 3 y^{2} z + 3 x^{2} y + 3 y z^{2} + 6 x yz + x^{3} + 3 x z^{2} + 3 x^{2} z = x^{3}

x^{3}

を左側に移動します

3 x^{2} y + 3 x^{2} z + 3 x y^{2} + 3 x z^{2} + 6 x yz + 3 y z^{2} + 3 y^{2} z = 0

標準的な形式で書く

a x^{2} + b x + c = 0

(3 y + 3 z) x^{2} + (3 y^{2} + 3 z^{2} + 6 yz) x + 3 y z^{2} + 3 y^{2} z = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) \pm ( 3 y ^{2} + 3 z ^{2} + 6 yz ) ^{2} - 4 ( 3 y + 3 z ) ( 3 y z ^{2} + 3 y ^{2} z )}{2 ( 3 y + 3 z )}

簡素化

(3 y^{2} + 3 z^{2} + 6 yz)^{2} - 4 (3 y + 3 z) (3 y z^{2} + 3 y^{2} z) : 3 (y + z) (y - z)

x_{1, 2} = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) \pm 3 ( y + z ) ( y - z )}{2 ( 3 y + 3 z )}; 3 y + 3 z \neq = 0

解を分離する

x_{1} = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) + 3 ( y + z ) ( y - z )}{2 ( 3 y + 3 z )}, x_{2} = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) - 3 ( y + z ) ( y - z )}{2 ( 3 y + 3 z )}

x = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) + 3 ( y + z ) ( y - z )}{2 ( 3 y + 3 z )} : - z

x = \frac{- ( 3 y ^{2} + 3 z ^{2} + 6 yz ) - 3 ( y + z ) ( y - z )}{2 ( 3 y + 3 z )} : - y

二次equationの解:

x = - z, x = - y; 3 y + 3 z \neq = 0

グラフ

人気の例

2x^0-13x^2+26x-24=0

2 x^{0} - 13 x^{2} + 26 x - 24 = 0

(d - x)^{2} = 2 x^{2}

(a+b)^2x^2+(a-b)^2=0

(a + b)^{2} x^{2} + (a - b)^{2} = 0

solvefor x,maxz=6x^1+4x^2

so l v e f or x, ma x z = 6 x^{1} + 4 x^{2}

solvefor x,f(x-2)=x^2+15

so l v e f or x, f (x - 2) = x^{2} + 15