find the product of (3x-5y-4)(9x^2+25y^2+15xy+12x-20y+16) .

Answers (1)

solution: we have , 

                (3x-5y-4)(9x^2+25y^2+15xy+12x-20y+16)

               =   left  3x+(-5y)+(-4) 
ight \left  (3x)^2+(-5y)^2+(-4)^2-(3x)(-5y)-(-5y)(-4)-3x(-4)
ight

                =   (a+b+c)(a^2+b^2+c^2-ab-bc-ca)  , where a=3x ,b=-5y ,c=-4

                    =a^3+b^3+c^3-3abc 

                  =(3x)^3+(-5y)^3+(-4)^3-3(3x)(-5y)(-4)=27x^3-125y^3-180xy-64 

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions