I just assumed that the cube root denests. Suppose ∛(55 + 63√2) = a + b√2. Cubing both sides, and equating "like terms" (with and without √2) 55 + 63√2 = (a + b√2)³ = a³ + 3a²(b√2) + 3a(b√2)² + (b√2)³ = a³ + 3√2a²b + 6ab² + 2√2b³ = (a³ + 6ab²) + (3a²b + 2b³)√2 So a³ + 6ab² = 55 & 3a²b + 2b³ = 63 a(a² + 6b²) = 55 & b(3a² + 2b²) = 63 Since this isn't supposed to be a ridiculous problem, I'll assume that a & b are integers. So the above two statements are then two products of integers to make 55 & 63. 55 is 1 * 55 or 55 = 5 * 11 63 is 1 * 63 or 3 * 21 or 7 * 9 Since the two equations involve squares & such, a & b can only each be 1 or 3. Let's try b = 1. Then, from the 2nd eqn (1) [ 3a² + 2(1)²] = 63 a² = 61/3 => a is not integer Let's try a = 1. Into 1st equation: (1) [(1)² + 6b²] = 55 b² = 9 => b = +/- 3 Checking (a,b) = (1,3) into 2nd eqn (1) [ (1)² + 6(3)² ] = 1 + 54 = 55 checks! [Clearly b = -3 isn't going to work. Discard] Check: (1 + 3√2)³ = 1³ + 3(1)²(3√2) + 3(1)(3√2)² + (3√2)³ = 1 + 9√2 + 54 + 54√2 = 55 + 63√2 So our answer is 1 + 3√2. Done!
I'm glad you recognized a similar question from a previous video! 💯💕🤩Thanks for noticing the connection! It's good to see people revisiting old problems. 🙏💡💯
There are better ways to denest cube roots of quadratic surds which do not require you to solve a quadratic equation, only a single cubic equation. To denest ∛(55 + 63√2), assume there exist _rational_ numbers a and b with √b _irrational_ such that (1) ∛(55 + 63√2) = a + √b Then, we must also have (2) ∛(55 − 63√2) = a − √b From (1) and (2) we have a² − b = (a + √b)(a − √b) = ∛(55 + 63√2)·∛(55 − 63√2) = ∛((55 + 63√2)·(55 − 63√2)) = ∛(55² − 63²·2) = ∛(3025 − 7938) = ∛(−4913) = −17, so (3) a² − b = −17 In case you are wondering: denestability requires a² − b = ∛(−4913) to be an integer, and since 10³ = 1000 and 20³ = 8000 this means that a² − b must be an integer between −20 and −10 with a cube −4913. But only integers ending in 7 have a cube ending in 3, which makes −17 the only candidate. Also from (1) the cube of a + √b must equal 55 + 63√2 so we have (a + √b)³ = a³ + 3a²√b + 3ab + b√b = (a³ + 3ab) + (3a² + b)√b = 55 + 63√2 which implies (4) a³ + 3ab = 55 From (3) we have b = a² + 17 and substituting this in (4) we get a³ + 3a(a² + 17) = 55 which gives (5) 4a³ + 51a − 55 = 0 Since a must be rational, we are looking for rational solutions of this cubic equation. In this case, it is easy to see that a = 1 is a solution of (5) since the sum of the coefficients of this cubic equation is zero. This is also the _only_ real root and _a fortiori_ the only rational root of (5) since the left hand side of (5) is strictly increasing on the real number line. With a = 1 we have b = a² + 17 = 18 and so (1) implies that we have ∛(55 + 63√2) = 1 + √18 which can also be written as ∛(55 + 63√2) = 1 + 3√2 Of course (2) implies that we also have ∛(55 − 63√2) = 1 − 3√2
I think that the surd in the solution must be sqrt2 ==#2 So 55 +63#2 = (x-y#2)^3 = xxx-3xxy#2 +6xyy -2#2yyy = x(xx+6yy) - #2.y(3xx+2yy)=55 +63#2 x(xx+6yy)=55 so x must divide 55, 5 is too big so x=1, hence y=3 check in the other equation.. y(3xx+2yy)==3(3+18)=63 ok alternatively use y(3xx+2yy) =63 y must divide 63 y=1,3,7,9.. try y=1 , 3xx+2=63 X impossible try y=3 3xx+18=21, x=1 substitute in the other equation to check.
Nice solution
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@superacademy247 thanks 👍
Интересно следить за Вашим решением. Спасибо
I just assumed that the cube root denests.
Suppose ∛(55 + 63√2) = a + b√2.
Cubing both sides, and equating "like terms" (with and without √2)
55 + 63√2 = (a + b√2)³
= a³ + 3a²(b√2) + 3a(b√2)² + (b√2)³
= a³ + 3√2a²b + 6ab² + 2√2b³
= (a³ + 6ab²) + (3a²b + 2b³)√2
So a³ + 6ab² = 55 & 3a²b + 2b³ = 63
a(a² + 6b²) = 55 & b(3a² + 2b²) = 63
Since this isn't supposed to be a ridiculous problem, I'll assume that a & b are integers.
So the above two statements are then two products of integers to make 55 & 63.
55 is 1 * 55 or 55 = 5 * 11
63 is 1 * 63 or 3 * 21 or 7 * 9
Since the two equations involve squares & such, a & b can only each be 1 or 3.
Let's try b = 1. Then, from the 2nd eqn
(1) [ 3a² + 2(1)²] = 63
a² = 61/3 => a is not integer
Let's try a = 1. Into 1st equation:
(1) [(1)² + 6b²] = 55
b² = 9 => b = +/- 3
Checking (a,b) = (1,3) into 2nd eqn
(1) [ (1)² + 6(3)² ] = 1 + 54 = 55 checks!
[Clearly b = -3 isn't going to work. Discard]
Check:
(1 + 3√2)³ = 1³ + 3(1)²(3√2) + 3(1)(3√2)² + (3√2)³
= 1 + 9√2 + 54 + 54√2
= 55 + 63√2
So our answer is 1 + 3√2. Done!
Thanks for sharing your your analysis 💕💯🤩🙏
Surd[55+63Sqrt[2],3]=1+3Sqrt[2] It’s in my head.
x=₃√(55+63√2) =p+q√2
x³=55+63√2 = ppp +3ppq√2+6pqq+2√2qqq
p(pp+6qq)=55
since integer p=1 q=3
q√2( 3pp+2qq)=63√2
3(3.1+2.9) =63
x=1+3√2
The same question solved on 22/6/2024 on your channel
I'm glad you recognized a similar question from a previous video! 💯💕🤩Thanks for noticing the connection! It's good to see people revisiting old problems. 🙏💡💯
There are better ways to denest cube roots of quadratic surds which do not require you to solve a quadratic equation, only a single cubic equation.
To denest ∛(55 + 63√2), assume there exist _rational_ numbers a and b with √b _irrational_ such that
(1) ∛(55 + 63√2) = a + √b
Then, we must also have
(2) ∛(55 − 63√2) = a − √b
From (1) and (2) we have a² − b = (a + √b)(a − √b) = ∛(55 + 63√2)·∛(55 − 63√2) = ∛((55 + 63√2)·(55 − 63√2)) = ∛(55² − 63²·2) = ∛(3025 − 7938) = ∛(−4913) = −17, so
(3) a² − b = −17
In case you are wondering: denestability requires a² − b = ∛(−4913) to be an integer, and since 10³ = 1000 and 20³ = 8000 this means that a² − b must be an integer between −20 and −10 with a cube −4913. But only integers ending in 7 have a cube ending in 3, which makes −17 the only candidate.
Also from (1) the cube of a + √b must equal 55 + 63√2 so we have (a + √b)³ = a³ + 3a²√b + 3ab + b√b = (a³ + 3ab) + (3a² + b)√b = 55 + 63√2 which implies
(4) a³ + 3ab = 55
From (3) we have b = a² + 17 and substituting this in (4) we get
a³ + 3a(a² + 17) = 55
which gives
(5) 4a³ + 51a − 55 = 0
Since a must be rational, we are looking for rational solutions of this cubic equation. In this case, it is easy to see that a = 1 is a solution of (5) since the sum of the coefficients of this cubic equation is zero. This is also the _only_ real root and _a fortiori_ the only rational root of (5) since the left hand side of (5) is strictly increasing on the real number line.
With a = 1 we have b = a² + 17 = 18 and so (1) implies that we have
∛(55 + 63√2) = 1 + √18
which can also be written as
∛(55 + 63√2) = 1 + 3√2
Of course (2) implies that we also have
∛(55 − 63√2) = 1 − 3√2
Thanks for your insightful demonstration 👌💕😍😎
I think that the surd in the solution must be sqrt2 ==#2
So 55 +63#2 = (x-y#2)^3 =
xxx-3xxy#2 +6xyy -2#2yyy = x(xx+6yy) - #2.y(3xx+2yy)=55 +63#2
x(xx+6yy)=55 so x must divide 55, 5 is too big so x=1, hence y=3
check in the other equation.. y(3xx+2yy)==3(3+18)=63 ok
alternatively use y(3xx+2yy) =63
y must divide 63 y=1,3,7,9..
try y=1 , 3xx+2=63 X impossible
try y=3 3xx+18=21, x=1
substitute in the other equation to check.