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If-ax-2-bx-c-0-had-two-roots-p-and-q-and-p-2-q-2-p-3-q-3-then-show-that-b-3-2a-2-c-ab-2-3abc-




Question Number 205353 by MATHEMATICSAM last updated on 17/Mar/24
If ax^2  + bx + c = 0 had two roots p and q  and p^2  + q^2  = p^3  + q^3  then show that  b^3  − 2a^2 c + ab^2  = 3abc.
$$\mathrm{If}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{had}\:\mathrm{two}\:\mathrm{roots}\:{p}\:\mathrm{and}\:{q} \\ $$$$\mathrm{and}\:{p}^{\mathrm{2}} \:+\:{q}^{\mathrm{2}} \:=\:{p}^{\mathrm{3}} \:+\:{q}^{\mathrm{3}} \:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{3}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {c}\:+\:{ab}^{\mathrm{2}} \:=\:\mathrm{3}{abc}. \\ $$
Answered by sniper237 last updated on 17/Mar/24
p^2 +q^2 =(p+q)^2 −2pq=(b^2 /a^2 )−((2c)/a)  p^3 +q^3 =(p+q)^3 −3pq(p+q)=−(b^3 /a^3 )+((3bc)/a^2 )  ⇒^(×a^3 )  −b^3 +3abc=ab^2 −2ac
$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\left({p}+{q}\right)^{\mathrm{3}} −\mathrm{3}{pq}\left({p}+{q}\right)=−\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }+\frac{\mathrm{3}{bc}}{{a}^{\mathrm{2}} } \\ $$$$\overset{×{a}^{\mathrm{3}} } {\Rightarrow}\:−{b}^{\mathrm{3}} +\mathrm{3}{abc}={ab}^{\mathrm{2}} −\mathrm{2}{ac} \\ $$
Answered by A5T last updated on 17/Mar/24
p+q=((−b)/a),pq=(c/a)  (p+q)^2 −2pq=(p+q)^3 −3pq(p+q)  ⇒(b^2 /a^2 )−((2c)/a)=((−b^3 )/a^3 )+((3bc)/a^2 )⇒ab^2 −2a^2 c=−b^3 +3abc
$${p}+{q}=\frac{−{b}}{{a}},{pq}=\frac{{c}}{{a}} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}=\left({p}+{q}\right)^{\mathrm{3}} −\mathrm{3}{pq}\left({p}+{q}\right) \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}=\frac{−{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }+\frac{\mathrm{3}{bc}}{{a}^{\mathrm{2}} }\Rightarrow{ab}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {c}=−{b}^{\mathrm{3}} +\mathrm{3}{abc} \\ $$

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