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If (314)4−(413)4(314)2−(413)2=(13a12)2\dfrac{\Big(3\dfrac{1}{4}\Big)^4 - \Big(4\dfrac{1}{3}\Big)^4}{\Big(3\dfrac{1}{4}\Big)^2 - \Big(4\dfrac{1}{3}\Big)^2} = \Big(\dfrac{13a}{12}\Big)^2, find a.

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(314)4−(413)4(314)2−(413)2=(13a12)2⇒(134)4−(133)4(134)2−(133)2=(13a12)2⇒((134)2−(133)2)((134)2+(133)2)((134)2−(133)2)=(13a12)2⇒(134)2+(133)2=(13a12)2⇒16916+1699=(13a12)2⇒169(116+19)=(13a12)2⇒169(9+16144)=(13a12)2⇒169(25144)=(13a12)2⇒(25×169144)=(13a12)2⇒(13×512)2=(13a12)2\dfrac{\Big(3\dfrac{1}{4}\Big)^4 - \Big(4\dfrac{1}{3}\Big)^4}{\Big(3\dfrac{1}{4}\Big)^2 - \Big(4\dfrac{1}{3}\Big)^2} = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \dfrac{\Big(\dfrac{13}{4}\Big)^4 - \Big(\dfrac{13}{3}\Big)^4}{\Big(\dfrac{13}{4}\Big)^2 - \Big(\dfrac{13}{3}\Big)^2} = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \dfrac{\Big(\Big(\dfrac{13}{4}\Big)^2 - \Big(\dfrac{13}{3}\Big)^2\Big)\Big(\Big(\dfrac{13}{4}\Big)^2 + \Big(\dfrac{13}{3}\Big)^2\Big)}{\Big(\Big(\dfrac{13}{4}\Big)^2 - \Big(\dfrac{13}{3}\Big)^2\Big)} = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \Big(\dfrac{13}{4}\Big)^2 + \Big(\dfrac{13}{3}\Big)^2 = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \dfrac{169}{16} + \dfrac{169}{9} = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ 169\Big(\dfrac{1}{16} + \dfrac{1}{9}\Big) = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ 169\Big(\dfrac{9 + 16}{144}\Big) = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ 169\Big(\dfrac{25}{144}\Big) = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \Big(\dfrac{25 \times 169}{144}\Big) = \Big(\dfrac{13a}{12}\Big)^2\\[1em] ⇒ \Big(\dfrac{13 \times 5}{12}\Big)^2 = \Big(\dfrac{13a}{12}\Big)^2\\[1em]

Hence, the value of a = 5.

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