\(S^o=\)\(\angle\)BAF + \(\angle\)ABC + \(\angle\)BCD + \(\angle\)CDE + \(\angle\)DEF + \(\angle\)EFA

  =\(\angle\)AFB + \(\angle\)BAF + \(\angle\)ABF + \(\angle\)CBF + \(\angle\)BCF + \(\angle\)DCF + \(\angle\)CDF + \(\angle\)EDF + \(\angle\)DEF + \(\angle\)EFD + \(\angle\)CFD + \(\angle\)BFC

 =\(180^o+180^o+180^o+180^o=720^o\)

Sory, I mistyped. It should've been \(720^o\) not \(540^o.\)

OMG !I don't khow what do you say!

I don't know I'm sorry but i'm grade 4 =)

We have:

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

Other way:

\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

From (*) and (**) we have  \(a+b\le2\)

Your ex is so hard to do :) 

We have:

a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)

Other way:

a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2

=(a+12b)2+34b2=(a+12b)2+34b2

Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b

34b2≥034b2≥0 with ∀b∀b

So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b

or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)

From (*) and (**) we have  a+b≤2a+b≤2

Your ex is so hard to do :) 

With x=2 then \(4P\left(2\right)=5.2^2=20\Rightarrow P\left(2\right)=5\)

With x=3 then \(P\left(3\right)+3P\left(2\right)=5.3^2=45\Rightarrow P\left(3\right)=45-5.3=30\)

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{zx}{80}=\dfrac{xy+yz+zx}{24+30+80}=\dfrac{1206}{134}=9\)

\(\Rightarrow x^2=64.9=8^2.3^2\Rightarrow x=\pm24\)

With x=24 then y=24:8.3=9;z=30

With x=-24 then y=-9;z=-30

We have:

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}\)

Apply the same sequence properties, we have:

\(\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}=\dfrac{xy+yz+xz}{24+30+80}=\dfrac{1206}{134}=9\)

So \(x^2=9.64=576\)\(\Rightarrow x=24\)

\(\dfrac{xy}{24}=9\Rightarrow xy=24.9=216\Rightarrow y=9\)

\(\dfrac{yz}{30}=9\Rightarrow yz=270\Rightarrow z=30\)

So \(\left(x;y;z\right)=\left(24;9;30\right)\)