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Given α , β are the roots of x²-p(x+1)-q=0 Now we have to find the value of (α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) But from equation it is clear that Sum of the roots = α+β = +p Product of the roots = α*β = -p-q (α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q (α+β)² = α²+β²+2αβ ⇒α²+β² = (α+β)² - 2αβ = p² - 2*(-p-q) = p²+2p+2q (1/α²+1/β²) = (α²+β²)/(αβ)² = (p²+2p+2q)/(-p-q)² = (p²+2p+2q)/(p²+q²+2pq) (α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q = p²+2p+2q+4p+ (p²+2p+2q)/(p²+q²+2pq)+2q = p²+6p+4q+ (p²+2p+2q)/(p²+q²+2pq) = {(p²+6p+4q)(p²+q²+2pq)+(p²+2p+2q)}/(p²+q²+2pq) = {p⁴+6p³+4qp²+p²q²+6pq²+4q³+2p³q+12p²q+8pq²+p²+2p+2q}/(p²+q²+2pq) This is the answer.