RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities Exercise 4.5

Exercise 4.5

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Question 1:

Find the following products:

(i) (3x + 2y + 2z) (9x² + 4y² + 4z² − 6xy − 4yz − 6zx)

(ii) (4x − 3y + 2z) (16x² + 9y² + 4z² + 12xy + 6yz − 8zx)

(iii) (2ab − 3b − 2c) (4a² + 9b² +4c² + 6 ab − 6 bc + 4ca)

(iv) (3x − 4y + 5z) (9x² +16y² + 25z² + 12xy −15zx + 20yz)

Answer 1:

In the given problem, we have to find Product of equations

(i)Given

We shall use the identity 

Hence the product of is

(ii) Given

We shall use the identity 

Hence the product of is

(iii) Given

We shall use the identity 

Hence the product of is

(iv) Given

We shall use the identity 

Hence the product of is

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Question 2:

Evaluate:

(i) 25³ − 75³ + 50³

(ii) 48³ − 30³ − 18³

(iii) ${\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^3$

(iv) (0.2)³ − (0.3)³ + (0.1)³

Answer 2:

In the given problem we have to evaluate the following

(i) Given 

We shall use the identity 

Let Take 

Hence the value of is

(ii) Given 

We shall use the identity 

Let Take 

Hence the value of is

(iii) Given 

We shall use the identity 

Let Take 

Applying least common multiple we get,

Hence the value of is

(iv) Given 

We shall use the identity 

Let Take 

$a^3+b^3+c^3=\left(0.2-0.3+0.1\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc$
$a^3+b^3+c^3=0\times \left(a^2+b^2+c^2-ab-bc-ca\right)+3abc$

Hence the value of is

Question 3:

If x + y + z = 8 and xy +yz +zx = 20, find the value of x³ + y³ + z³ −3xyz

Answer 3:

In the given problem, we have to find value of

Given

We shall use the identity

We know that 

Here substituting we get

Hence the value of is .

Question 4:

If a + b + c = 9 and ab +bc + ca = 26, find the value of a³ + b³+ c³ − 3abc

Answer 4:

In the given problem, we have to find value of

Given

We shall use the identity

We know that 

Here substituting we get, 

Hence the value of is .

Question 5:

If a + b + c = 9 and a²+ b² + c² =35, find the value of a³ + b³ + c³ −3abc

Answer 5:

In the given problem, we have to find value of

Given

We shall use the identity

We know that 

Here substituting we get

Hence the value of is .

Question 1:

Mark the correct alternative in each of the following:

(1) If $x+\frac{1}{x}=5$, then $x^2+\frac{1}{x^2}=$

(a) 25

(b) 10

(c) 23

(d) 27

Answer 1:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Hence the value of is

Hence the correct choice is (c).

Question 2:

If $x+\frac{1}{x}=2$, then $x^3+\frac{1}{x^3} =$

(a) 64

(b) 14

(c) 8

(d) 2

Answer 2:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (d).

Question 3:

If $x+\frac{1}{x}$ = 4, then $x^4+\frac{1}{x^4}=$

(a) 196

(b) 194

(c) 192

(d) 190

Answer 3:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Squaring on both sides we get, 

Hence the value of is

Hence the correct choice is (b).

Question 4:

If $x+\frac{1}{x}=3$, then $x^6+\frac{1}{x^6}$ =

(a) 927

(b) 414

(c) 364

(d) 322

Answer 4:

In the given problem, we have to find the value of

Given

We shall use the identityand

Here put,

Take Cube on both sides we get,

Hence the value of is

Hence the correct choice is (d).

Question 5:

 If $x^2+\frac{1}{x^2} =102$, then $x-\frac{1}{x}$ =

(a) 8

(b) 10

(c) 12

(d) 13

Answer 5:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (b).

Question 6:

If $x^3+\frac{1}{x^3}=110$, then $x+\frac{1}{x}=$

(a) 5

(b) 10

(c) 15

(d) none of these

Answer 6:

In the given problem, we have to find the value of

Given

We shall use the identity

Put we get,

Substitute y = 5 in the above equation we get

The Equation satisfy the condition that

Hence the value of is 5

The correct choice is (a).