Algebra Questions and Answers

l +m= -n

= (l + m)³ = (-n)³

= l³+m³+3lm (l+ m) + n³= 0

= l³+m³+3lm(-n) + n³= 0

Hence l³+ m³ + n³= 3lmn

a –  1/a =2

Squaring both the sides,

= (a – 1/a)²  = 4

= a² + 1/a² - 2 = 4

Or, (a²+1/a²) = 6

Squaring both the sides again,

(a²+1/a²)² = (6)²

(a⁴+1/a⁴+2) = 36

Therefore,

a⁴+1/a⁴= 34

First, let's square the given equation:

\left ( \frac{b-1}{b} \right )^{2} = 3^{2}

Expanding the left side of the equation:

b^{2} - 2(b)\frac{1}{b} + \left ( \frac{1}{b} \right )^{2} = 9

Simplifying further:

b^{2} - 2 + \left ( \frac{1}{b} \right )^{2} = 9

Now, let's rearrange the terms:

b^{2} + \left ( \frac{1}{b} \right )^{2} = 9 + 2

b^{2} + \left ( \frac{1}{b} \right )^{2} = 11

Now, let's square this equation again:

\left (b^{2} + \left ( \frac{1}{b} \right )^{2} \right )^{2} = 11^{2}

Expanding the left side:

b^{4} + 2(b^{2}) \left (\frac{1}{b^{2}} \right ) + \left ( \frac{1}{b} \right )^{4} = 121

Simplifying further:

b^{4} + 2 + \left ( \frac{1}{b} \right )^{4} = 121

Now, let's rearrange the terms:

b^{4} +  \left ( \frac{1}{b} \right )^{4} = 121- 2

b^{4} + \left ( \frac{1}{b} \right )^{4} = 119

Now, we can multiply the equation b^2 + 1/b^2 = 11 by the equation b^4 + 1/b^4 = 119:

(b^2 + 1/b^2)(b^4 + 1/b^4) = 11 * 119

Expanding both sides:

b^6 + (b^2)(1/b^4) + (1/b^2)(b^4) + 1/b^6 = 1309

Simplifying further:

b^6 + b^2/b^4 + b^4/b^2 + 1/b^6 = 1309

b^6 + 1/b^2 + b^2 + 1/b^6 = 1309

Now, we can substitute the value of b^2 + 1/b^2 from the previous equation (b^2 + 1/b^2 = 11):

b^6 + 11 + 1/b^6 = 1309

Simplifying further:

b^6 + 1/b^6 = 1309 - 11

b^6 + 1/b^6 = 1298

Therefore, the value of b^6 + 1/b^6 is 1298.

Given that f(p)= P² – 28p+196

Or,

P² – 28p + 196= 0

= (p-14)(p-14)= 0

Therefore p= 14

P₁m+ q₁n= 1

P₂m- q₂n= 1

In order to get a unique solution we know  p₁/p₂≠ q₁/-q₂

Now by looking at the options, it is evident that since the value of p and q cannot be 0, so option b and c gets eliminated.

On factorizing the equation ab -3a + 5b + r we get

= a(b-3)+5(b-3)= ab – 3a + 5b - 15

Therefore r= -15

Since Adam is a twice as good worker as Smith hence

His work is done per day 2x

And Smith work is done per day= x

Hence, Adam+Smith’s 1 day work= x+2x= 1/14

3x= 1/14

Therefore,

x= 1/14*3= 1/42 days

Hence Smith’s work done per day= 1/42

Adam’s efficiency is twice that of Smith, so time taken by him will be half = 42/2 =21

Hence Adam takes 21 days to finish the task, and Smith will take 2*21= 42 days to finish the same task if Adam is ill.

If we mark a daughter’s current age as x.

Then, her mother’s age at the time of her daughter’s birth will be,

(36-x) = x

2x = 36

X= 18.

Therefore, if the daughter is 18 years old at present then her age four years back will be,

18 - 4 = 14 years

Let the days be X

30*7*18 = 21*8*X

3780= 168X

Therefore,

X= 3780/168= 22.5 days

Let the number of balls be b

Then,

450/b – 450/b+5= 15

= 1/b – 1/(b+5) = 15/450

= (b+5-b)/b(b+5)= 1/30

= b² + 5b -150 = 0

= (b+15) (b-10) = 0

Therefore,

b= 10