If a=8+3root 7,b=1/a, find the value of a2+b2
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a= 8+∛7
b=1/a
⇒b = 1/8+∛7 × 8-∛7/8-∛7
⇒b = 8-∛7/(8)² - (∛7)²
⇒b = 8-∛7/64-63
⇒b = 8-∛7
thus
a²+b² =(8+∛7)² + (8-∛7)²
⇒a²+b² = 64 +63+48√7 +64+63 -48√7
⇒a² +b² =254
b=1/a
⇒b = 1/8+∛7 × 8-∛7/8-∛7
⇒b = 8-∛7/(8)² - (∛7)²
⇒b = 8-∛7/64-63
⇒b = 8-∛7
thus
a²+b² =(8+∛7)² + (8-∛7)²
⇒a²+b² = 64 +63+48√7 +64+63 -48√7
⇒a² +b² =254
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i also want to know the same questuion but the answer given below is incorrect as its square root not cube root
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