Answer :
Certainly! Let's prove that [tex]\( x^3 + 3bx = 2a \)[/tex] given [tex]\( x = \left(a + \sqrt{a^2 + b^3}\right)^{1/3} + \left(a - \sqrt{a^2 + b^3}\right)^{1/3} \)[/tex].
### Step-by-Step Solution:
1. Define the Variables:
Let [tex]\( m = \left(a + \sqrt{a^2 + b^3}\right)^{1/3} \)[/tex] and [tex]\( n = \left(a - \sqrt{a^2 + b^3}\right)^{1/3} \)[/tex].
2. Express [tex]\( x \)[/tex] in Terms of [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
Therefore, we have [tex]\( x = m + n \)[/tex].
3. Cube [tex]\( x \)[/tex]:
We are asked to find [tex]\( x^3 \)[/tex]. Using the binomial theorem for [tex]\( (m+n)^3 \)[/tex], we get:
[tex]\[ x^3 = (m + n)^3 = m^3 + n^3 + 3mn(m + n) \][/tex]
4. Compute [tex]\( m^3 \)[/tex] and [tex]\( n^3 \)[/tex]:
From the definitions, [tex]\( m^3 = a + \sqrt{a^2 + b^3} \)[/tex] and [tex]\( n^3 = a - \sqrt{a^2 + b^3} \)[/tex].
5. Sum [tex]\( m^3 \)[/tex] and [tex]\( n^3 \)[/tex]:
[tex]\[ m^3 + n^3 = (a + \sqrt{a^2 + b^3}) + (a - \sqrt{a^2 + b^3}) = 2a \][/tex]
6. Determine [tex]\( mn \)[/tex]:
To compute [tex]\( mn \)[/tex]:
[tex]\[ mn = \left( \left(a + \sqrt{a^2 + b^3}\right)^{1/3} \right) \left( \left(a - \sqrt{a^2 + b^3}\right)^{1/3} \right) \][/tex]
Notice:
[tex]\[ mn = \left( (a + \sqrt{a^2 + b^3})(a - \sqrt{a^2 + b^3}) \right)^{1/3} = \left( a^2 - (\sqrt{a^2 + b^3})^2 \right)^{1/3} = \left( a^2 - (a^2 + b^3) \right)^{1/3} = (-b^3)^{1/3} = -b \][/tex]
7. Substitute Values in the Expansion:
Recall the expansion [tex]\( x^3 = m^3 + n^3 + 3mn(m + n) \)[/tex]:
Using [tex]\( m+n = x \)[/tex], [tex]\( m^3 + n^3 = 2a \)[/tex], and [tex]\( mn = -b \)[/tex], it follows that:
[tex]\[ x^3 = 2a + 3(-b)x \][/tex]
8. Result:
Rearrange the terms:
[tex]\[ x^3 + 3bx = 2a \][/tex]
Thus, we have proved that:
[tex]\[ x^3 + 3bx = 2a \][/tex]
as required.
### Step-by-Step Solution:
1. Define the Variables:
Let [tex]\( m = \left(a + \sqrt{a^2 + b^3}\right)^{1/3} \)[/tex] and [tex]\( n = \left(a - \sqrt{a^2 + b^3}\right)^{1/3} \)[/tex].
2. Express [tex]\( x \)[/tex] in Terms of [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
Therefore, we have [tex]\( x = m + n \)[/tex].
3. Cube [tex]\( x \)[/tex]:
We are asked to find [tex]\( x^3 \)[/tex]. Using the binomial theorem for [tex]\( (m+n)^3 \)[/tex], we get:
[tex]\[ x^3 = (m + n)^3 = m^3 + n^3 + 3mn(m + n) \][/tex]
4. Compute [tex]\( m^3 \)[/tex] and [tex]\( n^3 \)[/tex]:
From the definitions, [tex]\( m^3 = a + \sqrt{a^2 + b^3} \)[/tex] and [tex]\( n^3 = a - \sqrt{a^2 + b^3} \)[/tex].
5. Sum [tex]\( m^3 \)[/tex] and [tex]\( n^3 \)[/tex]:
[tex]\[ m^3 + n^3 = (a + \sqrt{a^2 + b^3}) + (a - \sqrt{a^2 + b^3}) = 2a \][/tex]
6. Determine [tex]\( mn \)[/tex]:
To compute [tex]\( mn \)[/tex]:
[tex]\[ mn = \left( \left(a + \sqrt{a^2 + b^3}\right)^{1/3} \right) \left( \left(a - \sqrt{a^2 + b^3}\right)^{1/3} \right) \][/tex]
Notice:
[tex]\[ mn = \left( (a + \sqrt{a^2 + b^3})(a - \sqrt{a^2 + b^3}) \right)^{1/3} = \left( a^2 - (\sqrt{a^2 + b^3})^2 \right)^{1/3} = \left( a^2 - (a^2 + b^3) \right)^{1/3} = (-b^3)^{1/3} = -b \][/tex]
7. Substitute Values in the Expansion:
Recall the expansion [tex]\( x^3 = m^3 + n^3 + 3mn(m + n) \)[/tex]:
Using [tex]\( m+n = x \)[/tex], [tex]\( m^3 + n^3 = 2a \)[/tex], and [tex]\( mn = -b \)[/tex], it follows that:
[tex]\[ x^3 = 2a + 3(-b)x \][/tex]
8. Result:
Rearrange the terms:
[tex]\[ x^3 + 3bx = 2a \][/tex]
Thus, we have proved that:
[tex]\[ x^3 + 3bx = 2a \][/tex]
as required.
Answer:
Certainly! Let's prove that
3
+
3
=
2
x
3
+3bx=2a given
=
(
+
2
+
3
)
1
/
3
+
(
−
2
+
3
)
1
/
3
x=(a+
a
2
+b
3
)
1/3
+(a−
a
2
+b
3
)
1/3
.
### Step-by-Step Solution:
1. Define the Variables:
Let
=
(
+
2
+
3
)
1
/
3
m=(a+
a
2
+b
3
)
1/3
and
=
(
−
2
+
3
)
1
/
3
n=(a−
a
2
+b
3
)
1/3
.
2. Express
x in Terms of
m and
n :
Therefore, we have
=
+
x=m+n .
3. Cube
x :
We are asked to find
3
x
3
. Using the binomial theorem for
(
+
)
3
(m+n)
3
, we get:
3
=
(
+
)
3
=
3
+
3
+
3
(
+
)
x
3
=(m+n)
3
=m
3
+n
3
+3mn(m+n)
4. Compute
3
m
3
and
3
n
3
:
From the definitions,
3
=
+
2
+
3
m
3
=a+
a
2
+b
3
and
3
=
−
2
+
3
n
3
=a−
a
2
+b
3
.
5. Sum
3
m
3
and
3
n
3
:
3
+
3
=
(
+
2
+
3
)
+
(
−
2
+
3
)
=
2
m
3
+n
3
=(a+
a
2
+b
3
)+(a−
a
2
+b
3
)=2a
6. Determine
mn :
To compute
mn :
=
(
(
+
2
+
3
)
1
/
3
)
(
(
−
2
+
3
)
1
/
3
)
mn=((a+
a
2
+b
3
)
1/3
)((a−
a
2
+b
3
)
1/3
)
Notice:
=
(
(
+
2
+
3
)
(
−
2
+
3
)
)
1
/
3
=
(
2
−
(
2
+
3
)
2
)
1
/
3
=
(
2
−
(
2
+
3
)
)
1
/
3
=
(
−
3
)
1
/
3
=
−
mn=((a+
a
2
+b
3
)(a−
a
2
+b
3
))
1/3
=(a
2
−(
a
2
+b
3
)
2
)
1/3
=(a
2
−(a
2
+b
3
))
1/3
=(−b
3
)
1/3
=−b
7. Substitute Values in the Expansion:
Recall the expansion
3
=
3
+
3
+
3
(
+
)
x
3
=m
3
+n
3
+3mn(m+n) :
Using
+
=
m+n=x ,
3
+
3
=
2
m
3
+n
3
=2a , and
=
−
mn=−b , it follows that:
3
=
2
+
3
(
−
)
x
3
=2a+3(−b)x
8. Result:
Rearrange the terms:
3
+
3
=
2
x
3
+3bx=2a
Thus, we have proved that:
3
+
3
=
2
x
3
+3bx=2a
as required.
