Answer:
Step-by-step explanation:
According to question
α−d+α+α+d=−3ba
3α=−3ba
α=−ba
α2−dd+α2+αd+α2−d2=3a
3α2−d2=3ca
3b2a2−h2=3ca
d2=3b2a2−3ca=3b2−3aca2
α(α2−d2)=−da
−ba(b2a2−(3b2−3aca2))=−da
b(b2−3b2+3aca2)=d
−2b3+3ab=a2d
2b3−3abc+a2d=0
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