Solve {ccc}{a}&{b}&{c}{a-b}&{b-c}&{c-a}{b+c}&{c+a}&{a+b}=a^3+b^3+c^3-3ab

\left| \begin{array} { c c c } { a } & { b } & { c } \\ { a - b } & { b - c } & { c - a } \\ { b + c } & { c + a } & { a + b } \end{array} \right| = a ^ { 3 } + b ^ { 3 } + c ^ { 3 } - 3 a b

Evaluate

a^{3}-3abc+b^{3}+c^{3}=a^{3}-3ab+b^{3}+c^{3}

Quiz

5 problems similar to:

\left| \begin{array} { c c c } { a } & { b } & { c } \\ { a - b } & { b - c } & { c - a } \\ { b + c } & { c + a } & { a + b } \end{array} \right| = a ^ { 3 } + b ^ { 3 } + c ^ { 3 } - 3 a b

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Prove that the paraboloids: \displaystyle\frac{{x}^{{2}}}{{{a}_{{1}}^{{2}}}}+\frac{{y}^{{2}}}{{{b}_{{1}}^{{2}}}}=\frac{{{2}{z}}}{{c}_{{1}}} ; \displaystyle\frac{{x}^{{2}}}{{{a}_{{2}}^{{2}}}}+\frac{{y}^{{2}}}{{{b}_{{2}}^{{2}}}}=\frac{{{2}{z}}}{{c}_{{2}}} ...

https://socratic.org/questions/prove-that-the-paraboloids-x-2-a-1-2-y-2-b-1-2-2z-c-1-x-2-a-2-2-y-2-b-2-2-2z-c-2-1

Cesareo R. May 12, 2017 See https://socratic.org/questions/prove-that-the-paraboloids-x-2-a-1-2-y-2-b-1-2-2z-c-1-x-2-a-2-2-y-2-b-2-2-2z-c-2

Proof using properties of determinants

https://math.stackexchange.com/questions/718006/proof-using-properties-of-determinants

You have to compute D=\left| \begin{array}{ccc} y+z & y & z \\ x & x+z & z \\ x & y & x+y \end{array} \right| where x=a^2,y=b^2,z=c^2. You can proceed as follows : \begin{array}{lcl} D &=& \frac{1}{x}\left| \begin{array}{ccc} xy+xz & xy & xz \\ x & x+z & z \\ x & y & x+y \end{array} \right| \\ &=& \frac{1}{x}\left| \begin{array}{ccc} 0 & xy-y(x+z)-z(y) & xz-y(z)-z(x+y) \\ x & x+z & z \\ x & y & x+y \end{array} \right| \\ &=& -2yz\left| \begin{array}{ccc} 0 & 1 & 1 \\ 1 & x+z & z \\ 1 & y & x+y \end{array} \right| \\ \end{array} ...

An application of the mountain pass lemma

https://math.stackexchange.com/questions/755547/an-application-of-the-mountain-pass-lemma

You should set up the problem in the space W^{1,2}(]0,\pi[). Moreover the functional should be: \varphi(u):=\int_0^\pi \left(\frac12|u'|^2-\lambda u^2-\frac{u^4}{4}\right)dt which verifies the ...

Approximation By Simple Functions

https://math.stackexchange.com/questions/422195/approximation-by-simple-functions

Fix n. If t \geq n, it is clear that \phi_n(t)=n \leq t. And for t <n : let k2^{-n}\leq t<(k+1)2^{-n} for some k. Then \phi_n(t)=k2^{-n} \leq t. So, \phi_n(t)\leq t for any n. t ...

Variational formulation for bilaplacian problem

https://math.stackexchange.com/questions/144641/variational-formulation-for-bilaplacian-problem

See https://mathoverflow.net/questions/96870 for an explanation.

Logic behind cubic resolution of x^4+px^2+qx+r=0

https://math.stackexchange.com/questions/803659/logic-behind-cubic-resolution-of-x4px2qxr-0

This is only part of the way to a real answer, but I believe the point of the b_i is that they are linear polynomials in the subfield \mathbb{Q}(c_1,c_2,c_3,c_4)^{A_4} of the splitting field \mathbb{Q}(c_1,c_2,c_3,c_4) ...

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\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Simultaneous equation

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