Evaluate (a2b+2ba)2−(a2b−2ba)2−4\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4(2ba+a2b)2−(2ba−a2b)2−4
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Solving,
⇒(a2b+2ba)2−(a2b−2ba)2−4⇒(a2b)2+(2ba)2+2×a2b×2ba−[(a2b)2+(2ba)2−2×a2b×2ba]−4⇒a24b2+4b2a2+2−[a24b2+4b2a2−2]−4⇒a24b2+4b2a2+2−a24b2−4b2a2+2−4⇒4−4⇒0.\Rightarrow \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 + 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a} - \Big[\Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 - 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \Big[\dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} - 2\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a^2} + 2 - 4 \\[1em] \Rightarrow 4 - 4 \\[1em] \Rightarrow 0.⇒(2ba+a2b)2−(2ba−a2b)2−4⇒(2ba)2+(a2b)2+2×2ba×a2b−[(2ba)2+(a2b)2−2×2ba×a2b]−4⇒4b2a2+a24b2+2−[4b2a2+a24b2−2]−4⇒4b2a2+a24b2+2−4b2a2−a24b2+2−4⇒4−4⇒0.
Hence, (a2b+2ba)2−(a2b−2ba)2−4\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4(2ba+a2b)2−(2ba−a2b)2−4 = 0.
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