The given equation is equivalent to (x+7)³ - 8 = 8•2•³√(x+8) (*). Let ³√(x+8) = y (1) => x+8=y³ => x+7= y³-1 and the (*) rewritten (y³-1)³ - 8 = 16 y => y⁹ - 3y⁶+ 3y³ - 16y-9 =0 => y=-1 or y⁸-y⁷+x⁶-4y⁵+4y⁴-4y³+7y²-7y-9=0 (2) For y=-1 => -1 = ³√(x+8) due to (1). => (-1)³ = x+8 => x = -9 is satisfy the equation (*).
Let x=u-8. Then the equation becomes (u-i)^3-8=16u^1/3. So, 1/2[(u/2-1/2)^3-1]=u^1/3. Let a=1/2(u-1).Then, we get, upon cubing, a^9-3a^6+3a^3-16a-9=0 or, (a+1)(a^2-a-1)(a^6+2a^4-2a^3+4a^2-2a+9)=0. a^6+2a^4-2a^3+4a^2-2a+9 is always positive. So, a=-1 which gives x=-9 and a=1/2[1+/-√5], which gives x=-6+/-√5. All are valid solutions. x=-9, -6+/-√5.
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Another great problem! Thank you for sharing.
The given equation is equivalent to
(x+7)³ - 8 = 8•2•³√(x+8) (*).
Let ³√(x+8) = y (1) => x+8=y³ =>
x+7= y³-1 and the (*) rewritten
(y³-1)³ - 8 = 16 y =>
y⁹ - 3y⁶+ 3y³ - 16y-9 =0 =>
y=-1 or y⁸-y⁷+x⁶-4y⁵+4y⁴-4y³+7y²-7y-9=0 (2)
For y=-1 => -1 = ³√(x+8) due to (1).
=> (-1)³ = x+8 => x = -9 is satisfy the equation (*).
Let x=u-8. Then the equation becomes (u-i)^3-8=16u^1/3. So, 1/2[(u/2-1/2)^3-1]=u^1/3. Let a=1/2(u-1).Then, we get, upon cubing, a^9-3a^6+3a^3-16a-9=0 or, (a+1)(a^2-a-1)(a^6+2a^4-2a^3+4a^2-2a+9)=0. a^6+2a^4-2a^3+4a^2-2a+9 is always positive. So, a=-1 which gives x=-9 and a=1/2[1+/-√5], which gives x=-6+/-√5. All are valid solutions. x=-9, -6+/-√5.
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x=-9
x^3+21) ➖ 8 21x^3 ➖ ((8^2){21x^2 ➖ 64}=43x^2 43^1x^2 1^1x^2 1x^2 (x ➖ 2x+1).{x^3+x^3 ➖ }+{8x+8x ➖ }+64={x^6+16x^2}=16x^8+{64+64 ➖ }={16x^8+128}=144x^8 1^2^2^2^2x^2^3 1^1^1^1^1x^1^1^2 1x^2(x ➖ 2x+1).
Странные действия автора
Сначала он выводит 8 из под корня, потом наоборот вводит в корень. Зачем?
X=-9