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IF one root of x2 + px + q = 0 be square of the other, Then prove p³+ q² +q = 3 pq

$L e t α & β b e t h e r o o t s o f g i v e n p o l y n o m i a l x^{2} + p x + q = 0 t h e r e f o r e, α + β = - p & α β = q a l s o, α = β^{2} n o w, β^{2} + β = - p & β^{2} . β = q \Rightarrow β^{3} = q 3 p q = 3 (- β^{2} - β) β^{3} = - 3 β^{4} (β + 1) ............... (i) a n d p^{3} + q^{2} + q = - {(β^{2} + β)}^{3} + β^{6} + β^{3} = - β^{6} - β^{3} - 3 β^{4} . β - 3 β^{2} . β^{2} + β^{6} + β^{3} = - 3 β^{5} - 3 β^{4} = - 3 β^{4} (β + 1) ............... (i i) f r o m (i) & (i i), i t s p r o v e d p^{3} + q^{2} + q = 3 p q$

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