Dalla frase all espressione

(4/3·a - b)·(a + b)^2 - a^3/(a^2 - b^2)

a = - 1/2

b = 2

(4/3·(- 1/2) - 2)·(- 1/2 + 2)^2 - (- 1/2)^3/((- 1/2)^2 - 2^2)=

=(- 2/3 - 2)·(- 1/2 + 2)^2 - (- 1/2)^3/(1/4 - 2^2)=

=(- 8/3)·(3/2)^2 - (- 1/2)^3/(1/4 - 4)=

=(- 8/3)·(9/4) - (- 1/2)^3/(- 15/4)=

=-6 - (- 1/8)/(- 15/4)=

=-6 - 1/30=- 181/30

(4/3 a - b) * (a + b)^2  - a^3 / (a^2 - b^2)....

a = - 1/2

b = 2

[4/3 *(- 1/2) - 2] * (- 1/2 + 2)^2 - (- 1/2)^3 /[(- 1/2)^2 - 2^2] =

= [- 2/3 - 2] * (- 1/2 + 4/2)^2 - (- 1/8) / [1/4 - 4]  =

= [- 2/3 - 6/3] * (+ 3/2)^2 - (- 1/8) /[1/4 - 16/4] =

= - 8/3 * (+ 9/4) - (- 1/8) : (-15/4) =

= - 6  - (- 1/8) * (- 4/15) =

= - 6 -  (+ 1/30) =

= - 180/30 - 1/30 = - 181/30.

Ciao  @fabroxy

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$\small \left[\left(\dfrac{4}{3}a-b\right)·\left(a+b\right)^2\right]-\dfrac{a^3}{a^2-b^2}=$

sostituisci le incognite con i valori indicati:

$\small =\left[\left(\dfrac{4}{3}·-\dfrac{1}{2}-2\right)·\left(-\dfrac{1}{2}+2\right)^2\right]-\dfrac{\left(-\dfrac{1}{2}\right)^3}{\left(-\dfrac{1}{2}\right)^2-2^2}=$

$\small =\left[\left(\dfrac{\cancel4^2}{3}·-\dfrac{1}{\cancel2_1}-2\right)·\left(\dfrac{-1+4}{2}\right)^2\right]-\dfrac{-\dfrac{1}{8}}{\dfrac{1}{4}-4}=$

$\small =\left[\left(\dfrac{2}{3}·-\dfrac{1}{1}-2\right)·\left(\dfrac{3}{2}\right)^2\right]-\dfrac{-\dfrac{1}{8}}{\dfrac{1-16}{4}}=$

$\small =\left[\left(-\dfrac{2}{3}-2\right)·\dfrac{9}{4}\right]-\dfrac{-\dfrac{1}{8}}{-\dfrac{15}{4}}=$

$\small =\left[\left(\dfrac{-2-6}{3}\right)·\dfrac{9}{4}\right]-\left(-\dfrac{1}{8}\right)·\left(-\dfrac{4}{15}\right)=$

$\small =\left[-\dfrac{\cancel8^2}{\cancel3_1}·\dfrac{\cancel9^3}{\cancel4_1}\right]-\left(-\dfrac{1}{\cancel8_2}\right)·\left(-\dfrac{\cancel4^1}{15}\right)=$

$\small =\left[-\dfrac{2}{1}·\dfrac{3}{1}\right]-\left(-\dfrac{1}{2}\right)·\left(-\dfrac{1}{15}\right)=$

$\small =-2·3-\dfrac{1}{30}=$

$\small =-6-\dfrac{1}{30}=$

$\small =\dfrac{-180-1}{30}=$

$\small =-\dfrac{181}{30}=$