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Question Number 116854 by mnjuly1970 last updated on 07/Oct/20

$. . . c a l c u l u s e l e m e n t a r y a l g e b r a . . .$ $p l e a s e s o l v e ::$ $\sqrt[3]{6 x + 9} + \sqrt[3]{7 - 7 x} + \sqrt[3]{x - 8} = 2$ $. . . m . n . j u l y . 1970 . . .$
	Answered by MJS_new last updated on 07/Oct/20

	$easier than it seems$ $\sqrt[3]{6 x + 9} + \sqrt[3]{x - 8} = 2 + \sqrt[3]{7 x - 7}$ $let t = \sqrt[3]{7 x - 7} \Leftrightarrow x = \frac{t^{3} + 7}{7}$ $\frac{\sqrt[3]{6 t^{3} + 105} + \sqrt[3]{t^{3} - 49}}{\sqrt[3]{7}} = t + 2$ ${(\sqrt[3]{6 t^{3} + 105} + \sqrt[3]{t^{3} - 49})}^{3} = 7 {(t + 2)}^{3}$ $using {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b) we get$ $3 \sqrt[3]{7} (\sqrt[3]{6 t^{3} + 105} \times \sqrt[3]{t^{3} - 49}) (t + 2) = 42 t (t + 2)$ $\Rightarrow t_{1} = - 2 \Rightarrow x_{1} = - 1 / 7$ $the rest^{3} leads to$ $t^{6} - \frac{581}{6} t^{3} - \frac{1715}{2} = 0$ $t = \sqrt[3]{7 x - 7}$ $49 x^{2} - \frac{4655}{6} x - \frac{392}{3} = 0$ $\Rightarrow x_{2} = - 1 / 6 \land x_{3} = 16$
	Commented by bemath last updated on 07/Oct/20

	$greatt$
	Commented by mnjuly1970 last updated on 07/Oct/20

	$t h a n k y o u m r m . j . s . . .$
	Answered by TANMAY PANACEA last updated on 07/Oct/20
	$a^3 =6x+9 b^3 =7−7x c^3 =x−8 a^3 +b^3 +c^3 =8=2^3 a+b+c=(a^3 +b^3 +c^3 )^(1/3) (a+b)^3 +3(a+b)^2 c+3(a+b)c^2 +c^3 =a^3 +b^3 +c^3 3ab(a+b)+3(a+b)c(a+b+c)=0 3(a+b)(ab+ac+bc+c^2 )=0 3(a+b){a(b+c)+c(b+c)}=0 (a+b)(b+c)(a+c)=0 when a+b=0 a^3 =−b^3 6x+9=−(7−7x) 6x−7x=−16 x=16 when b+c=0 b^3 =−c^3 7−7x=−x+8 −6x=1 x=((−1)/6) when a+c=0 a^3 =−c^3 6x+9=−x+8 7x=−1 x=((−1)/7) x=16,((−1)/6) and ((−1)/7)$
	$a^{3} = 6 x + 9 b^{3} = 7 - 7 x c^{3} = x - 8$ $a^{3} + b^{3} + c^{3} = 8 = 2^{3}$ $a + b + c = {(a^{3} + b^{3} + c^{3})}^{\frac{1}{3}}$ ${(a + b)}^{3} + 3 {(a + b)}^{2} c + 3 (a + b) c^{2} + c^{3} = a^{3} + b^{3} + c^{3}$ $3 a b (a + b) + 3 (a + b) c (a + b + c) = 0$ $3 (a + b) (a b + a c + b c + c^{2}) = 0$ $3 (a + b) {a (b + c) + c (b + c)} = 0$ $(a + b) (b + c) (a + c) = 0$ $w h e n a + b = 0$ $a^{3} = - b^{3}$ $6 x + 9 = - (7 - 7 x)$ $6 x - 7 x = - 16 x = 16$ $w h e n b + c = 0$ $b^{3} = - c^{3}$ $7 - 7 x = - x + 8$ $- 6 x = 1 x = \frac{- 1}{6}$ $w h e n a + c = 0$ $a^{3} = - c^{3}$ $6 x + 9 = - x + 8$ $7 x = - 1 x = \frac{- 1}{7}$ $x = 16, \frac{- 1}{6} a n d \frac{- 1}{7}$
	Commented by mnjuly1970 last updated on 07/Oct/20

	$t h a n k y o u s o m u c h m r$ $t a n m a y . .$
	Commented by TANMAY PANACEA last updated on 07/Oct/20

	$m o s t w e l c o m e s i r$
	Answered by 1549442205PVT last updated on 07/Oct/20
	$Put a=((6x+9))^(1/3) ,b=((7−7x))^(1/3) ,c=((x−8))^(1/3) We have a^3 +b^3 +c^3 =8,a+b+c=2 Hence,applying the identity : a^3 +b^3 +c^3 =(a+b+c)^3 −3(a+b)(b+c)(c+a) we get:8=8−3(a+b)(b+c)(c+a) ⇒(a+b)(a+c)(b+c)=0 i)a+b=0⇔^3 (√(6x+9))=−^3 (√(7−7x)) ⇔6x+9=−(7−7x)⇔x=16 ii)a+c=0⇔^3 (√(6x+9))=−^3 (√(x−8)) ⇔6x+9=−(x−8)⇔x=−1/7 iii)b+c=0⇔^3 (√(7−7x))=−^3 (√(x−8)) ⇔7−7x=−(x−8)⇔x=−1/6 Thus,the given equation has three roots: x∈{16,−1/7;−1/6}$
	$Put a = \sqrt[3]{6 x + 9}, b = \sqrt[3]{7 - 7 x}, c = \sqrt[3]{x - 8}$ $We have a^{3} + b^{3} + c^{3} = 8, a + b + c = 2$ $Hence, applying the identity :$ $a^{3} + b^{3} + c^{3} = {(a + b + c)}^{3} - 3 (a + b) (b + c) (c + a)$ $we get : 8 = 8 - 3 (a + b) (b + c) (c + a)$ $\Rightarrow (a + b) (a + c) (b + c) = 0$ $i) a + b = 0 \Leftrightarrow^{3} \sqrt{6 x + 9} = -^{3} \sqrt{7 - 7 x}$ $\Leftrightarrow 6 x + 9 = - (7 - 7 x) \Leftrightarrow x = 16$ $ii) a + c = 0 \Leftrightarrow^{3} \sqrt{6 x + 9} = -^{3} \sqrt{x - 8}$ $\Leftrightarrow 6 x + 9 = - (x - 8) \Leftrightarrow x = - 1 / 7$ $iii) b + c = 0 \Leftrightarrow^{3} \sqrt{7 - 7 x} = -^{3} \sqrt{x - 8}$ $\Leftrightarrow 7 - 7 x = - (x - 8) \Leftrightarrow x = - 1 / 6$ $Thus, the given equation ha s three$ $roots : x \in {16, - 1 / 7; - 1 / 6}$
	Commented by mnjuly1970 last updated on 07/Oct/20

	$v e r y n i c e . . .$
	Commented by 1549442205PVT last updated on 31/Oct/20

	$You are welcome$
	Answered by mnjuly1970 last updated on 08/Oct/20

	$w e k n o w t h a t :$ $a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$ $a = \sqrt[3]{6 x + 9}, b = \sqrt[3]{7 - 7 x}, c = \sqrt[3]{x - 8} - 2$ $6 x + 9 + 7 - 7 x + x - 8 - 8 - 6 \sqrt[3]{{(x - 8)}^{2}} + 12 \sqrt[3]{x - 8}$ $= 3 (\sqrt[3]{6 x + 9}) (\sqrt[3]{7 - 7 x}) (\sqrt[3]{x - 8} - 2)$ $- 6 \sqrt[3]{x - 8} (\sqrt[3]{x - 8} - 2) = 3 (\sqrt[3]{6 x + 9)} . (\sqrt[3]{7 - 7 x)} . (\sqrt[3]{x - 8} - 2)$ $x = 16$ $- 8 (x - 8) = (6 x + 9) (7 - 7 x)$ $- 8 x + 64 = 42 x - 42 x^{2} - 63 x + 63$ $42 x^{2} + 13 x + 1 = 0$ $Δ = 169 - 168 = 1$ $x = \frac{- 13 \underline{+} 1}{84} = \frac{- 14}{84} = \frac{- 1}{6}, \frac{- 1}{7}$ $. . . m . n . 1970 . . .$
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