Question

Heya!!
solve this....✌️❣️
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Answer 1

1.

Using remainder theorem, ...

This question can be solved by putting x=1/2

2.

Without actually calculating the cubes, ...

$A^3+B^3+C^3=3ABC$

'SUM OF CUBES EQUALS MULTIPLICATION'

$3\times 16\times (-7)\times (-9)=+3024$

3.

Simplify : $\sqrt{1^3+2^3+3^3}$

$\sqrt{1+8+27} =\sqrt{36} =6$

Answer 2

1. As per remainder theorum.

g(x) = 0

2x-1 = 0

x = 1/2

2. As we know about an identity in the algebra

i.e. a³ + b³ + c³ = 3abc

(16)³ + (-7)³ + (-9)³ = 3(16)(-7)(-9)

(16)³ + (-7)³ + (-9)³ = 3024

3. Expression :- (1³+2³+3³)½

= (1³+2³+3³)½

= (1+8+27)½

= (36)½

= (6)²×½

= 6

Ahh! Thanks for the question.