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Mathematics

What is the simplified form of the expression

b4−a4a(a+b)−b3ab2−ab+a2\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}?

Factorisation

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Answer

Given,

b4−a4a(a+b)−b3ab2−ab+a2\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}

Simplifying the numerator,

⇒b4−a4a(a+b)−b3a⇒(b2)2−(a2)2a(a+b)−b3a⇒(b2−a2)(b2+a2)a(a+b)−b3a⇒(b−a)(b+a)(b2+a2)a(a+b)−b3a⇒(b−a)(b2+a2)a−b3a⇒b3+ba2−ab2−a3a−b3a⇒b3+ba2−ab2−a3−b3a⇒ba2−ab2−a3a⇒a(ba−b2−a2)a⇒(ab−b2−a2)⇒−(a2+b2−ab).\Rightarrow \dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2)^2 - (a^2)^2}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2 - a^2)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b + a)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b^2 + a^2)}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3 - b^3}{a} \\[1em] \Rightarrow \dfrac{ba^2 - ab^2 - a^3}{a} \\[1em] \Rightarrow \dfrac{a(ba - b^2 - a^2)}{a} \\[1em] \Rightarrow (ab - b^2 - a^2) \\[1em] \Rightarrow -(a^2 + b^2 - ab).

Substituting value of numerator in given fraction,

⇒−(a2+b2−ab)b2−ab+a2⇒−(a2+b2−ab)(a2+b2−ab)⇒−1.\Rightarrow \dfrac{-(a^2 + b^2 - ab)}{b^2 - ab + a^2} \\[1em] \Rightarrow \dfrac{-(a^2 + b^2 - ab)}{(a^2 + b^2 - ab)} \\[1em] \Rightarrow -1.

Hence, b4−a4a(a+b)−b3ab2−ab+a2=−1\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2} = -1.

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