**a Cube plus b Cube Formula :**

In this section, we are going to see the formula for

a^{3 }+ b^{3}

We already know the formula/expansion for (a + b)^{3}.

That is,

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

**Case 1 : **

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Subtract 3ab(a + b) from each side.

(a + b)^{3} - 3ab(a + b) = a^{3} + b^{3}

Therefore, the formula for (a^{3} + b^{3}) is

**a ^{3} + b^{3 }= (a + b)^{3} - 3ab(a + b)**

**Case 2 : **

From case 1,

a^{3} + b^{3 }= (a + b)^{3} - 3ab(a + b)

a^{3} + b^{3}^{ }= (a + b)[(a + b)^{2} - 3ab]

a^{3} + b^{3 }= (a + b)[a^{2} + 2ab + b^{2 }- 3ab]

a^{3} + b^{3 }= (a + b)(a^{2} - ab + b^{2})

Therefore, the formula for (a^{3} + b^{3}) is

**a ^{3} + b^{3 }= (a + b)(a^{2} - ab + b^{2})**

So,

(a + b) and (a^{2} - ab + b^{2})

are the factors of (a^{3} + b^{3}).

**Note : **

Based on our need, either we can use the formula in case 1 or in case 2 for (a^{3} + b^{3}).

**Question 1 :**

Factor :

x^{3} + 8

**Solution :**

**Write (x ^{3} + 8) in the form of (a^{3} + b^{3}).**

x^{3} + 8 = x^{3} + 2^{3}

(x^{3} + 2^{3})^{ }is in the form of (a^{3 }+ b^{3}).

Comparing (a^{3 }+ b^{3})^{ }and (x^{3 }+ 2^{3}), we get

a = x

b = 2

Write the formula for (a^{3 }+ b^{3}) given in case 2 above.

a^{3} + b^{3 }= (a + b)(a^{2} - ab + b^{2})

Substitute x for a and 2 for b.

x^{3} + 2^{3} = (x + 2)(x^{2} - 2x + 2^{2})

x^{3} + 8 = (x + 2)(x^{2} - 2x + 4)

**Question 2 :**

Factor :

27x^{3} + 64

**Solution :**

**Write (27x ^{3} + 64) in the form of (a^{3} + b^{3}).**

27x^{3} + 64 = (3x)^{3} + 4^{3}

(3x)^{3} + 4^{3}^{ }is in the form of (a^{3 }+ b^{3}).

Comparing (a^{3 }+ b^{3})^{ }and (3x)^{3 }+ 4^{3}, we get

a = 3x

b = 4

Write the formula for (a^{3 }+ b^{3}) given in case 2 above.

a^{3} + b^{3 }= (a + b)(a^{2} - ab + b^{2})

Substitute 3x for a and 4 for b.

(3x)^{3} + 4^{3} = (3x + 4)[(3x)^{2} - (3x)(4) + 4^{2}]

27x^{3} + 64 = (3x + 4)(9x^{2} - 12x + 16)

**Question 3 :**

Factor :

8x^{3} + 27y^{3}

**Solution :**

**Write (8x ^{3} + **27y

8x^{3} + 27y^{3} = (2x)^{3} + (3y)^{3}

(2x)^{3} + (3y)^{3}^{ }is in the form of (a^{3 }+ b^{3}).

Comparing (a^{3 }+ b^{3})^{ }and (2x)^{3 }+ (3y)^{3}, we get

a = 2x

b = 3y

Write the formula for (a^{3 }+ b^{3}) given in case 2 above.

a^{3} + b^{3 }= (a + b)(a^{2} - ab + b^{2})

Substitute 2x for a and 3y for b.

(2x)^{3} + (3y)^{3} = (2x + 3y)[(2x)^{2} - (2x)(3y) + (3y)^{2}]

8x^{3} + 27y^{3} = (2x + 3y)(8x^{2} - 6xy + 9y^{2})

**Question 4 :**

Find the value of (x^{3} + y^{3}), if x + y = 4 and xy = 5.

**Solution :**

Write (x^{3 }+ y^{3}) in terms of (x + y) and xy using the formula given in case 1 above.

x^{3} + y^{3 }= (x + y)^{3} - 3xy(x + y)

Substitute 4 for (x + y) and 5 for xy.

x^{3} + y^{3 }= (4)^{3} - 3(5)(4)

x^{3} + y^{3 }= 64 - 60

x^{3} + y^{3 }= 4

After having gone through the stuff above, we hope that the students would have understood the formula for a cube plus b cube.

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