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given; sin⁴/a+cos⁴/b=1/a+b

prove that sin⁸/a³+cos⁸/b³=1/(a+b)³

$G i v e n, \frac{\sin^{4} a}{a} + \frac{\cos^{4} a}{b} = \frac{1}{a + b} ...... (1) \Rightarrow \frac{b \sin^{4} a + a \cos^{4} a}{a b} = \frac{1}{a + b} \Rightarrow (a + b) (b \sin^{4} a + a \cos^{4} a) = a b \Rightarrow (a b \sin^{4} a + a^{2} \cos^{4} a + b^{2} \sin^{4} a + a b \cos^{4} a) - a b = 0 \Rightarrow a b (\sin^{4} a + \cos^{4} a - 1) + a^{2} \cos^{4} a + b^{2} \sin^{4} a = 0 \Rightarrow a b ({(\sin^{2} a + \cos^{2} a)}^{2} - 2 \sin^{2} a \cos^{2} a - 1) + a^{2} \cos^{4} a + b^{2} \sin^{4} a = 0 \Rightarrow a b ({(1)}^{2} - 2 \sin^{2} a \cos^{2} a - 1) + a^{2} \cos^{4} a + b^{2} \sin^{4} a = 0 \Rightarrow a b (- 2 \sin^{2} a \cos^{2} a) + a^{2} \cos^{4} a + b^{2} \sin^{4} a = 0 \Rightarrow a^{2} \cos^{4} a + b^{2} \sin^{4} a - 2 a b \sin^{2} a \cos^{2} a = 0 \Rightarrow {(b \sin^{2} a - a \cos^{2} a)}^{2} = 0 \Rightarrow b \sin^{2} a - a \cos^{2} a = 0 \Rightarrow b \sin^{2} a = a \cos^{2} a \Rightarrow \frac{\sin^{2} a}{a} = \frac{\cos^{2} a}{b} ...... (2) F r o m (1), \sin^{2} a \frac{\sin^{2} a}{a} + \cos^{2} a \frac{\cos^{2} a}{b} = \frac{1}{a + b} \Rightarrow \sin^{2} a \frac{\sin^{2} a}{a} + \cos^{2} a \frac{\sin^{2} a}{a} = \frac{1}{a + b} (f r o m (2)) \Rightarrow \frac{\sin^{2} a}{a} (\sin^{2} a + \cos^{2} a) = \frac{1}{a + b} \Rightarrow \frac{\sin^{2} a}{a} = \frac{1}{a + b} S i m i l a r l y, \frac{\cos^{2} a}{b} = \frac{1}{a + b} T h u s, \frac{\sin^{2} a}{a} = \frac{\cos^{2} a}{b} = \frac{1}{a + b} ..... (3) N o w, \frac{\sin^{8} a}{a^{3}} + \frac{\cos^{8} a}{b^{3}} = \sin^{2} a {(\frac{\sin^{2} a}{a})}^{3} + \cos^{2} a {(\frac{\cos^{2} a}{b})}^{3} = \sin^{2} a {(\frac{1}{a + b})}^{3} + \cos^{2} a {(\frac{1}{a + b})}^{3} (f r o m (3)) = {(\frac{1}{a + b})}^{3} (\sin^{2} a + \cos^{2} a) = {(\frac{1}{a + b})}^{3} H e n c e, \frac{\sin^{8} a}{a^{3}} + \frac{\cos^{8} a}{b^{3}} = {(\frac{1}{a + b})}^{3}$

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