Question

Factorize : 125p3+ 27q3. Also, mention the identity used here

Answer 1

Basis

• Identity
• Expansion and Factorization

Let's expand a product.

$(a+b)(a^2-ab+b^2)$

$=a(a^2-ab+b^2)+b(a^2-ab+b^2)$

$=a^3-a^2b+ab^2+a^2b-ab^2+b^3$

$=a^3+b^3$

BODMAS doesn't change the result of the calculation. So, the equation is always true, which means an identity.

Now we can say the expansion is

$(a+b)(a^2-ab+b^2)=a^3+b^3$

Expansion is the reverse of factorization. So, the sum of the cubes can be factorized.

Solution

$125p^3$ is a perfect cube of $5p$. $27q^3$ is a perfect cube of $3q$.

So, it's a sum of two cubes.

Using identity,

$(5p)^3+(3q)^3=(5p+3q)(25p^2-15pq+9p^2)$

Answer 2

Basis

Identity

Expansion and Factorization

Let's expand a product.

(a+b)(a^2-ab+b^2)(a+b)(a

2

−ab+b

2

)

=a(a^2-ab+b^2)+b(a^2-ab+b^2)=a(a

2

−ab+b

2

)+b(a

2

−ab+b

2

)

=a^3-a^2b+ab^2+a^2b-ab^2+b^3=a

3

−a

2

b+ab

2

+a

2

b−ab

2

+b

3

=a^3+b^3=a

3

+b

3

BODMAS doesn't change the result of the calculation. So, the equation is always true, which means an identity.

Now we can say the expansion is

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

2

−ab+b

2

)=a

3

+b

3

Expansion is the reverse of factorization. So, the sum of the cubes can be factorized.

Solution

125p^3125p

3

is a perfect cube of 5p5p . 27q^327q

3

is a perfect cube of 3q3q .

So, it's a sum of two cubes.

Using identity,

(5p)^3+(3q)^3=(5p+3q)(25p^2-15pq+9p^2)(5p)

3

+(3q)

3

=(5p+3q)(25p

2

−15pq+9p

2