Question

if a and b are the zeros of polynomial x2-ax+a,find value of -1/a3+1/b3​

Answer 1

Answer:

Given :

• 'a' and 'b' are the zeros

• of the polynomial x² - ax +a

To find :

  value of 1/b³ - 1/a³

Solving Question :

         first we have to simplify the question and then substitute the value in the formulas that we know.

Formulas :

α + β = -b/a

α*β = c/a

Solutions:

   x³ - ax+a

-b = -a

⇒  b = a

c = a

a = 1

then, to simplify the question

$\dfrac{1}{b^3}-\dfrac{1}{a^3} =\dfrac{a^3-b^3}{(ab)^3}\\\\\implies \dfrac{(a-b)(a^2+ab+b^2)}{(ab)^3}\\\\then,\:a^2+b^2=(a+b)^2-2ab\\\\substitute\\\\\implies \dfrac{(a-b)(ab+(a+b)^2-2ab}{(ab)^3}\\\\\implies \dfrac{(a-b)((a+b)^2-ab)}{(ab)^3}$

$(a-b)^2= a^2+b^2+2ab\\\\(a-b)^2= (a+b)^2-2ab\\\\(a-b)= \sqrt{(a+b)^2-2ab$

thus our question becomes

$\dfrac{(\sqrt{(a+b)^2-2ab)}*(a+b)^2-ab)}{(ab)^3}\\\\substitute\: values(a\:and\:b \:are\:zeros)\\\\\implies \dfrac{\sqrt{(\dfrac{-b}{a})^2-2\dfrac{c}{a}}*(\dfrac{-b}{a})^2-\dfrac{c}{a})}{(\dfrac{c}{a})^3}\\\\\implies\boxed{ \dfrac{(\sqrt{a^2-2a})*(a^2-a)}{a^3}}$