Question

find theproduct using suitable identity (a+3) (a-3 )( a square +9)

Answer 1

(a+3)(a-3)(a²+9)

= (a²-3²)(a²+9) Using (a+b)(a-b)=a²-b²

= (a²-9)(a²+9)

= ((a²)²-9²) Using (a+b)(a-b)=a²-b²

= (a²×²-81)

= (a raise to the power 4-81)

HOPE YOU ARE SATISFIED

Answer 2

Answer:

Main Aim:-

• To find the value of $\sf{(a + 3)(a - 3)( {a}^{2} + 9)}$

Algebraic Indentity to be used:-

• $\boxed{\tt\purple{ (x + y)(x - y) = {x}^{2} - {y}^{2} }}$

Real Content:

$\sf{(a + 3)(a - 3)( {a}^{2} + 9)}$

(Expression given in the question)

$\sf{=[ ({a}^{2} - {3}^{2})][ {a}^{2} + 9 ]}$

(In the above expression, we have recognised that (a+3)(a-3) = a²-3² as (x+y)(x-y) = x²-y²)

$\sf{ =({a}^{2} - {3}^{2})( {a}^{2} + {3}^{2} )}$

(Above it is written that 9 can also be written as 3•3 i.e 3²)

Now, as we have got the Expression in form of (x+y)(x-y), we will try to find it's factorised form. That is, x²-y² where,

• x = a²
• y = 3²

$\sf{=( {{a}^{2} )}^{2} - ( {{3}^{2} )}^{2}}$

(Now we have written half but still modifications are required.)

$\sf{= ({a}^{2 \times 2})-({3}^{2 \times 2)}}$

(Almost Done! Now we have to change something in place of 3^4 and a^2•2)

$\sf{=({a}^{4}-{3}^{4})}$(Exponents are multiplied)

$\sf\green{={a}^{4}-81}$

(As we know, 3⁴ = 81, hence we have placed the value)

$\text{(Product \ Form/Answer)}$

_________...

REQUIRED ANSWER:-

$\tt{\therefore}$ Product of $\boxed{\large\tt{(a + 3)(a - 3)( {a}^{2} + 9)}}$ is $\boxed{\huge\tt\green{{a}^{4}-81}}$.