Harrington Coleman
10/21/2023 · Middle School

1. Show that : \( (\sqrt{3}+\sqrt{2}) 3+(\sqrt{3}-\sqrt{2}) 3=18 \sqrt{3} \)

UpStudy ThothAI Solution

Tutor-Verified Answer

Step-by-step Solution

To show that \[ (\sqrt{3}+\sqrt{2}) 3 + (\sqrt{3}-\sqrt{2}) 3 = 18 \sqrt{3}, \] we can start by simplifying the left-hand side. First, we can factor out the common term \(3\): \[ 3 \left( (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) \right). \] Next, we simplify the expression inside the parentheses: \[ (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}. \] Now, substituting this back into our expression gives: \[ 3 \cdot 2\sqrt{3} = 6\sqrt{3}. \] However, we need to check if this equals \(18\sqrt{3}\). It seems there was a misunderstanding in the original equation. Let's re-evaluate the original expression: \[ (\sqrt{3}+\sqrt{2}) 3 + (\sqrt{3}-\sqrt{2}) 3 = 3(\sqrt{3}+\sqrt{2} + \sqrt{3}-\sqrt{2}) = 3(2\sqrt{3}) = 6\sqrt{3}. \] Thus, we find that: \[ 6\sqrt{3} \neq 18\sqrt{3}. \] Therefore, the statement \[ (\sqrt{3}+\sqrt{2}) 3 + (\sqrt{3}-\sqrt{2}) 3 = 18 \sqrt{3} \] is incorrect. The correct simplification shows that the left-hand side equals \(6\sqrt{3}\).

Quick Answer

The statement is incorrect. The left-hand side simplifies to \(6\sqrt{3}\), not \(18\sqrt{3}\).
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text

Enter your question here…

By image
Re-Upload
Uploaded Files
xxxx.png0%
Submit
📸 STUDY CAN BE A REAL STRUGGLE
Why Not UpStudy It?
Select your plan below
Premium

You can enjoy

  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to answer and
    solution
Basic
  • Limited Solutions