1. Se dau numerele complexe \( a,\ b,\ c,\ d \) cu proprietatea ca \( a+b+c+d=0 \). Sa se dem ca:
\( a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab) \).
2. Demonstrati ca expresia
\( 2(a^4+b^4+c^4+d^4)-( a^2+b^2+c^2+d^2)^2+8abcd \)
este divizibila cu \( a+b+c+d \), a, b, c, d din Z
folosind relatiile lui Viete.
a, b, c, d
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Marius Mainea
- Gauss
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1) Folosim relatia (Newton) \( a^3+b^3+c^3+d^3-(a^2+b^2+c^2+d^2)(a+b+c+d)+(a+b+c+d)(ab+ac+ad+bc+bd+cd)-3(abc+bcd+cda+dab)=0 \)
2) Deasemenea
\( a^4+b^4+c^4+d^4-(a^3+b^3+c^3+d^3)(a+b+c+d)+(a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd)-(a+b+c+d)(abc+bcd+cda+dab)+4abcd=0 \) sau
\( 2\sum{a^4}-2\sum{a^2}\sum{ab}+8abcd=\mathcal{M}(a+b+c+d) \) sau
\( 2\sum{a^4-\sum{a^2}((\sum a)^2-\sum{a^2})+8abcd=\mathcal{M}(a+b+c+d) \)
sau concluzia problemei.
2) Deasemenea
\( a^4+b^4+c^4+d^4-(a^3+b^3+c^3+d^3)(a+b+c+d)+(a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd)-(a+b+c+d)(abc+bcd+cda+dab)+4abcd=0 \) sau
\( 2\sum{a^4}-2\sum{a^2}\sum{ab}+8abcd=\mathcal{M}(a+b+c+d) \) sau
\( 2\sum{a^4-\sum{a^2}((\sum a)^2-\sum{a^2})+8abcd=\mathcal{M}(a+b+c+d) \)
sau concluzia problemei.