**a plus b Whole Cube Formula :**

In this section, we are going to see the formula/expansion for (a + b)^{3}.

That is,

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

Multiply (a + b) and (a + b).

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

Simplify.

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

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

Combine the like terms.

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

or

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

**Problem 1 :**

Expand :

(x + 2)^{3 }

**Solution :**

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

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

a = x

b = 2

Write the formula / expansion for (a + b)^{3}.

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

Substitute x for a and 2 for b.

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

(x + 2)^{3} = x^{3} + 6x^{2} + 3(x)(4) + 8

(x + 2)^{3} = x^{3} + 6x^{2} + 12x + 8

So, the expansion of (x + 2)^{3} is

x^{3} + 6x^{2} + 12x + 8

**Problem 2 :**

Expand :

(2x + 3)^{3 }

**Solution :**

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

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

a = 2x

b = 3

Write the formula / expansion for (a + b)^{3}.

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

Substitute 2x for a and 3 for b.

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

(2x + 3)^{3} = 8x^{3} + 3(4x^{2})(3) + 3(2x)(9) + 27

(2x + 3)^{3} = 8x^{3} + 36x^{2} + 54x + 27

So, the expansion of (2x + 3)^{3} is

8x^{3} + 36x^{2} + 54x + 27

**Problem 3 :**

Expand :

(p + 2q)^{3 }

**Solution :**

(p + 2q)^{3 }is in the form of (a + b)^{3}

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

a = p

b = 2q

Write the formula / expansion for (a + b)^{3}.

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

Substitute p for a and 2q for b.

(p + 2q)^{3} = p^{3} + 3(p^{2})(2q) + 3(p)(2q)^{2} + (2q)^{3}

(p + 2q)^{3} = p^{3} + 6p^{2}q + 3(p)(4q^{2}) + 8q^{3}

(p + 2q)^{3} = p^{3} + 6p^{2}q + 12pq^{2} + 8q^{3}

So, the expansion of (p + 2q)^{3} is

p^{3} + 6p^{2}q + 12pq^{2} + 8q^{3}

**Problem 4 : **

If a + b = 12 and a^{3} + b^{3} = 468, then find the value of ab.^{ }

**Solution :**

To find the value of ab, we can use the formula or expansion for (a + b)^{3}.

Write the formula / expansion for (a + b)^{3}.

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

or

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

Substitute 12 for (a + b) and 468 for (a^{3} + b^{3}).

(12)^{3} = 468 + 3(ab)(12)

Simplify.

1728 = 468 + 36ab

Subtract 468 from each side.

1260 = 36ab

Divide each side by 36.

35 = ab

So, the value of ab is 35.

**Problem 5 :**

Find the value of :

(107)^{3}

**Solution :**

We can use the algebraic formula for (a + b)^{3 }and find the value of (107)^{3 }easily.

Write (107)^{3 }in the form of (a + b)^{3}.

(107)^{3} = (100 + 7)^{3}

Write the expansion for (a + b)^{3}.

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

Substitute 100 for a and 7 for b.

(100 + 7)^{3} = 100^{3} + 7^{3} + 3(100)(7)(100 + 7)

(100 + 7)^{3} = 1000000 + 343 + 3(100)(7)(107)

(100 + 7)^{3} = 1000000 + 343 + 224700

(107)^{3} = 1225043

So, the value of (107)^{3} is

1,225,043

Apart from the stuff explained above, if you would like to learn about more identities in Algebra,

After having gone through the stuff given above, we hope that the students would have understood the formula or expansion for cube of a binomial.

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