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2009年2月12日 · 数学题:已知x+y+z=0,求证x的3次方+y的3次方+z的3次方=3xyz(x的3次方+y的3次方+z的3次方)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)因为x+y+z=0则(x的3次方+y的3次方+z的3次方)-3xyz=0(x的3次方+y的3次方+z的3次方)=3xyz

If x + y + z = 0, show that x³ + y³ + z³ = 3xyz. Solution: Using the identity: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx) If x + y + z = 0, then the entire RHS becomes 0. Hence the …

分解因式： \(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+3xyz\). 它如果能分解，那就有一次因式，而且还是其次轮换式。 可以验证一次因式是 \(x+y+z\) .

2012年11月21日 · x³+y³+z³-3xyz =x³+3x²y+3xy²+y³+z³-3x²y-3xy²-3xyz =(x+y)³+z³-3xy(x+y+z) =(x+y+z)(x²+2xy+y²-xz-yz-3xy) =(x+y+z)(x²+y²+z²-xy-yz-xz)

例1.已知 x,y,z>0 ，证明 x^3+y^3+z^3\geq 3xyz . 这是我们熟知的 三元均值不等式，下面给出证明. 那么 x^3+y^3+z^3\geq 3xyz ，当且仅当 x=y=z 时等号成立. 看完上面的解答，很自然的一个问题是上面的因式分解是怎么“注意到”的.接下来看具体过程. 第一步， 注意到 x+y+z 是 x^3+y^3+z^3-3xyz 的一个因式 (注意不到就真没办法了). 最后根据 (x^2+y^2+z^2-xy-yz-zx)=\frac {1} {2}\left ( (x-y)^2+ (y-z)^2+ (z-x)^2 \right) 代入上式就得到答案里的式子了.

由10， x+y+z=0\ 时，有 x^3+y^3+z^3=3xyz\ 注意因式分解前为0，分解后每个括号都可能等于零，尤其不要漏掉10右侧，虽然都是平方但也可以等于零 待定系数法

2010年2月13日 · x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx) =(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]/2 ≥0 x^3+y^3+z^3≥3xyz

2020年9月11日 · To solve for 'x' in the equation 3xyz = P, we need to isolate x. This can be done by dividing both sides of the equation by 3yz, which gives the solution as x = P / (3yz). Explanation: The question asks you to solve for x in the equation 3xyz = P. Here, x is one of the variables and . P is a constant.

2024年12月13日 · Ex 2.4, 12 - Verify that x3 + y3 + z3 - 3xyz = 1/2 (x + y + z) ... Last updated at Dec. 13, 2024 by Teachoo. Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo.