Esercizio$\left(2senx-3cosx^3\right)^4$
Soluzione passo-passo
Passi intermedi
1
Applicare la formula: $\left(a+b\right)^4$$=a^4+4a^3b+6a^2b^2+4ab^3+b^4$, dove $a=2\sin\left(x\right)$, $b=-3\cos\left(x\right)^3$ e $a+b=2\sin\left(x\right)-3\cos\left(x\right)^3$
$\left(2\sin\left(x\right)\right)^4-12\left(2\sin\left(x\right)\right)^3\cos\left(x\right)^3+6\left(2\sin\left(x\right)\right)^2\left(-3\cos\left(x\right)^3\right)^2+8\sin\left(x\right)\left(-3\cos\left(x\right)^3\right)^3+\left(-3\cos\left(x\right)^3\right)^4$
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Passi intermedi
2
Applicare la formula: $\left(ab\right)^n$$=a^nb^n$
$16\sin\left(x\right)^4-12\cdot 8\sin\left(x\right)^3\cos\left(x\right)^3+6\cdot 4\sin\left(x\right)^2\left(-3\cos\left(x\right)^3\right)^2+8\sin\left(x\right)\left(-3\cos\left(x\right)^3\right)^3+\left(-3\cos\left(x\right)^3\right)^4$
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3
Applicare la formula: $ab$$=ab$, dove $ab=-12\cdot 8\sin\left(x\right)^3\cos\left(x\right)^3$, $a=-12$ e $b=8$
$16\sin\left(x\right)^4-96\sin\left(x\right)^3\cos\left(x\right)^3+6\cdot 4\sin\left(x\right)^2\left(-3\cos\left(x\right)^3\right)^2+8\sin\left(x\right)\left(-3\cos\left(x\right)^3\right)^3+\left(-3\cos\left(x\right)^3\right)^4$
4
Applicare la formula: $ab$$=ab$, dove $ab=6\cdot 4\sin\left(x\right)^2\left(-3\cos\left(x\right)^3\right)^2$, $a=6$ e $b=4$
$16\sin\left(x\right)^4-96\sin\left(x\right)^3\cos\left(x\right)^3+24\sin\left(x\right)^2\left(-3\cos\left(x\right)^3\right)^2+8\sin\left(x\right)\left(-3\cos\left(x\right)^3\right)^3+\left(-3\cos\left(x\right)^3\right)^4$
Risposta finale al problema $16\sin\left(x\right)^4-96\sin\left(x\right)^3\cos\left(x\right)^3+24\sin\left(x\right)^2\left(-3\cos\left(x\right)^3\right)^2+8\sin\left(x\right)\left(-3\cos\left(x\right)^3\right)^3+\left(-3\cos\left(x\right)^3\right)^4$