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SOLUTION: (x+y+z)^3. Click here to see ALL problems on Exponents. Question 631629: (x+y+z)^3. Answer by lenny460 (1073) ( Show Source ): You can put this solution on YOUR website! (x+y+z)^3. (x + y + z) (x + y + z) (x + y + z) We multiply using the FOIL Method: x * x = x^2. x * y = xy. x * z = xz. y * x = xy. y * y = y^2. y * z = yz. z * x = xz.
Detailed step by step solution for expand (x+y+z)^3.
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x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)
Please check the expression entered or try another topic. (x+y+ z)3 ( x + y + z) 3. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.
Mar 11, 2021 · The equation x 3 +y 3 +z 3 =k is known as the sum of cubes problem. While seemingly straightforward, the equation becomes exponentially difficult to solve when framed as a “Diophantine equation” — a problem that stipulates that, for any value of k, the values for x, y, and z must each be whole numbers.
I think you mean x+y+z=3. The left inequality: \sum_{cyc}x\sqrt{1+y^3}\geq\sum_{cyc}x=3. The right inequality. Let \{x,y,z\}=\{a,b,c\}, where a\geq b\geq c. Thus, by AM-GM, Rearrangement and ...
