A similar problem solved by you was squaring (a^3+b^3)^2=65^2 so a^6+2(ab)^3+a^6=4225 . Also cubing (a^2+b^2)^3=17^3 so a^6+3*(ab)^2*(a^2+b^2)+b^6=4913. Then substructing them we get 2(ab)^3 -3(ab)^2*17=-688 so if p=ab then 2p^3-51p^2+688=0. Then exactly as you do.
I want to point out although this is a very good video there is a mistake close to the end. When rejecting the “t” values that are not integers we cannot just simply disregard it. The problem states that x must be an integer that does not imply that t which is equal to AB must also be an integer. Because A=sqrt(x+8) so if x is an integer it doesn’t mean A has to be an integer unless x+8 is a perfect square. But we don’t know that. Therefore you must consider the positive number that was not an integer. You can disregard case 3 with the negative number since A is positive and B is positive then it follows AB=t>0.
in every equation of this sort that I have seen can be solved by separating the rhs into the sum of powers seen on the left hand side. In this case we have a^(3/2) + b^(3/2) = 64 +1 = 16^(3/2) +1^(3/2) so the solution is a=16,b=1 and a=1, b=16 note with a=9-x, b=x+8 a+b=17 … a=16, x=-7, so b=1 or a=1, x=8, b=16
A similar problem solved by you was squaring (a^3+b^3)^2=65^2 so a^6+2(ab)^3+a^6=4225 . Also cubing (a^2+b^2)^3=17^3 so a^6+3*(ab)^2*(a^2+b^2)+b^6=4913. Then substructing them we get 2(ab)^3 -3(ab)^2*17=-688 so if p=ab then 2p^3-51p^2+688=0. Then exactly as you do.
I want to point out although this is a very good video there is a mistake close to the end. When rejecting the “t” values that are not integers we cannot just simply disregard it. The problem states that x must be an integer that does not imply that t which is equal to AB must also be an integer. Because A=sqrt(x+8) so if x is an integer it doesn’t mean A has to be an integer unless x+8 is a perfect square. But we don’t know that. Therefore you must consider the positive number that was not an integer. You can disregard case 3 with the negative number since A is positive and B is positive then it follows AB=t>0.
in every equation of this sort that I have seen can be solved by separating the rhs into the sum of powers seen on the left hand side.
In this case we have a^(3/2) + b^(3/2) = 64 +1 = 16^(3/2) +1^(3/2)
so the solution is a=16,b=1 and a=1, b=16
note with a=9-x, b=x+8 a+b=17
… a=16, x=-7, so b=1
or a=1, x=8, b=16
x=8,-7
X=-7, or 8.
X= 8; -7
Even though x is integer, a and b may not be integer, so t may not be integer.
Crack this Radical Equation: √[(x + 8)³] + √[(9 - x)³] = 65, x ϵ Z; x =?
x + 8 > 0, 9 - x > 0; 9 > x > - 8, x, positive or negative integer
65 > √[(x + 8)³] > √[(9 - x)³] or 65 > √[(9 - x)³] > √[(x + 8)³]
√[(x + 8)³] + √[(9 - x)³] = 65 = 4³ + 1 = (√16)³ + 1 = [√(8 + 8)]³ + √[(9 - 8)³]; x = 8
√[(x + 8)³] + √[(9 - x)³] = 65 = 1 + (√16)³ = [√(- 7 + 8)]³ + √[9 - (- 7)³]; x = - 7
Answer check:
x = 8 or x = - 7: √[(x + 8)³] + √[(9 - x)³] = 65; Confirmed as shown
Final answer:
x = 8 or x = - 7
(x^3+24)+(x^3 ➖ 729)=24x^3+{x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ }={24x^3+x^3}=24x^6 8^8^8x^6:2^3^2^3^2^3x^6 1^1^1^1^1^1^1^1^1x^2^3;x^2^3 (x ➖ 3x+2).