Esercizio
$\left(\frac{1}{x^3}\:-\:3x\right)^4$
Soluzione passo-passo
Passi intermedi
1
Applicare la formula: $\left(a+b\right)^4$$=a^4+4a^3b+6a^2b^2+4ab^3+b^4$, dove $a=\frac{1}{x^3}$, $b=-3x$ e $a+b=\frac{1}{x^3}-3x$
$\left(\frac{1}{x^3}\right)^4-12\left(\frac{1}{x^3}\right)^3x+6\left(\frac{1}{x^3}\right)^2\left(-3x\right)^2+\frac{4\left(-3x\right)^3}{x^3}+\left(-3x\right)^4$
Passi intermedi
2
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=x^3$ e $n=4$
$\frac{1}{x^{12}}-12\left(\frac{1}{x^3}\right)^3x+6\left(\frac{1}{x^3}\right)^2\left(-3x\right)^2+\frac{4\left(-3x\right)^3}{x^3}+\left(-3x\right)^4$
Passi intermedi
3
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=x^3$ e $n=3$
$\frac{1}{x^{12}}-12\left(\frac{1}{x^{9}}\right)x+6\left(\frac{1}{x^3}\right)^2\left(-3x\right)^2+\frac{4\left(-3x\right)^3}{x^3}+\left(-3x\right)^4$
Passi intermedi
4
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=x^3$ e $n=2$
$\frac{1}{x^{12}}-12\left(\frac{1}{x^{9}}\right)x+6\left(\frac{1}{x^{6}}\right)\left(-3x\right)^2+\frac{4\left(-3x\right)^3}{x^3}+\left(-3x\right)^4$
Risposta finale al problema
$\frac{1}{x^{12}}-12\left(\frac{1}{x^{9}}\right)x+6\left(\frac{1}{x^{6}}\right)\left(-3x\right)^2+\frac{4\left(-3x\right)^3}{x^3}+\left(-3x\right)^4$