(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....


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Question Number 186292 by ajfour last updated on 03/Feb/23
$(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....$
$(x - p) (x^{3} - x - \frac{1}{3}) = 0$ $x^{4} - {p x}^{3} - x^{2} + (p - \frac{1}{3}) x + \frac{p}{3} = 0$ $(x^{2} + a x + h) (x^{2} + b x + k) = 0$ $a + b = - p$ $h + k + a b = - 1$ $b h + a k = p - \frac{1}{3}$ $h k = \frac{p}{3}$ $- - - - - -$ $s a y a b = t$ $- - - - - -$ $a h + b k + p - \frac{1}{3} = p (1 + t)$ $b h + a k - (p - \frac{1}{3}) = 0$ $\Rightarrow (a - b) (h - k) = p t - p + \frac{2}{3}$ $s q u a r i n g$ $(p^{2} - 4 t) {{(1 + t)}^{2} - \frac{4 p}{3}} = {(p t - p + \frac{2}{3})}^{2}$ $s a y t + 1 = z$ $(p^{2} + 4 - 4 z) (z^{2} - \frac{4 p}{3}) = {(p z - 2 p + \frac{2}{3})}^{2}$ $\Rightarrow$ $- 4 z^{3} + (p^{2} + 4) z^{2} + \frac{16 p z}{3} - \frac{4 p}{3} (p^{2} + 4)$ $= p^{2} z^{2} - 4 p (p - \frac{1}{3}) z + 4 {(p - \frac{1}{3})}^{2}$ $\Rightarrow z^{3} - z^{2} - p (p + 1) z + {(p - \frac{1}{3})}^{2}$ $+ \frac{p}{3} (p^{2} + 4) = 0$ $. . . . .$
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