Math, asked by amitasinghgps2411, 9 months ago

factorise 27r3 - 3r5

Answers

Answered by rishu6845

Answer:

3 r³ ( 3 + r ) ( 3 - r )

Step-by-step explanation:

Given---> 27 r³ - 3r⁵

To find ---> Factors of given expression

Solution---> ATQ,

27 r³ - 3r⁵

First we do prime factorization of 27

27 = 3 × 3 × 3

Now returning to original problem ,

27 r³ - 3r⁵

= 3 × 3 × 3 r³ - 3 r⁵

= 3 ( 3 × 3 r³ - r⁵ )

= 3 ( 9 r³ - r⁵ )

We can write r⁵ as r³ × r² by applying law of exponent ( aᵐ aⁿ = aᵐ⁺ⁿ )

= 3 ( 9 r³ - r³ r² )

Now we take r³ common from the bracket,

= 3 r³ ( 9 - r² )

= 3 r³ { ( 3 )² - ( r )² ]

We have an identity , a² - b² = ( a + b ) ( a - b ) , applying it here , we get,

= 3 r³ ( 3 + r ) ( 3 - r )

#Answerwithquality & #BAL

Answered by Aɾꜱɦ

$<font color =“green”>$

${ \huge \bf{ \mid{ \overline{ \underline{Answer}}} \mid}}$

$<body bgcolor= “black” ><fontcolor=“black ”>$

#answerwithquality #bal

Previous Question

Next Question