Answer :
Sure! Let's simplify the expression [tex]\(\left(x^{a+2} - 2x^a + 3^{a+1}\right)\left(x^a + x^{a+1}\right)\)[/tex] step-by-step.
1. Write down the expression:
[tex]\[ \left(x^{a+2} - 2x^a + 3^{a+1}\right)\left(x^a + x^{a+1}\right) \][/tex]
2. Distribute each term in the first parenthesis to each term in the second parenthesis. This means we'll multiply each term in [tex]\((x^{a+2} - 2x^a + 3^{a+1})\)[/tex] by each term in [tex]\((x^a + x^{a+1})\)[/tex].
3. First, distribute [tex]\(x^{a+2}\)[/tex]:
[tex]\[ x^{a+2} \cdot x^a + x^{a+2} \cdot x^{a+1} = x^{(a+2)+a} + x^{(a+2)+(a+1)} = x^{2a+2} + x^{2a+3} \][/tex]
4. Next, distribute [tex]\(-2x^a\)[/tex]:
[tex]\[ -2x^a \cdot x^a + -2x^a \cdot x^{a+1} = -2x^{a+a} + -2x^a \cdot x^{a+1} = -2x^{2a} + -2x^{a+(a+1)} = -2x^{2a} - 2x^{2a+1} \][/tex]
5. Finally, distribute [tex]\(3^{a+1}\)[/tex]:
[tex]\[ 3^{a+1} \cdot x^a + 3^{a+1} \cdot x^{a+1} = 3^{a+1}x^a + 3^{a+1}x^{a+1} \][/tex]
6. Combine all the expanded terms:
[tex]\[ x^{2a+2} + x^{2a+3} - 2x^{2a} - 2x^{2a+1} + 3^{a+1} x^a + 3^{a+1}x^{a+1} \][/tex]
So, the simplified expression is:
[tex]\[ 33^ax^a + 33^ax^{(a+1)} - 2x^{2a} - 2x^ax^{(a+1)} + x^ax^{(a+2)} + x^{(a+1)}x^{(a+2)} \][/tex]
Rewriting it:
[tex]\[ 33^ax^a + 33^ax^{a + 1} - 2x^{2a} - 2x^{a}* x^{a + 1} + x^{a + 1 + a + 1} + x^{a + 2 + a + 1} \][/tex]
Therefore, the expanded form of the given expression [tex]\(\left(x^{a+2} - 2x^a + 3^{a+1}\right)\left(x^a + x^{a+1}\right)\)[/tex] is given by:
[tex]\[ 3 \cdot 3^a \cdot x^a + 3 \cdot 3^a \cdot x^{a+1} - 2 \cdot x^{2a} - 2 \cdot x^{a} \cdot x^{a+1} + x^{a} \cdot x^{a+2} + x^{a+1} \cdot x^{a+2} \][/tex]
1. Write down the expression:
[tex]\[ \left(x^{a+2} - 2x^a + 3^{a+1}\right)\left(x^a + x^{a+1}\right) \][/tex]
2. Distribute each term in the first parenthesis to each term in the second parenthesis. This means we'll multiply each term in [tex]\((x^{a+2} - 2x^a + 3^{a+1})\)[/tex] by each term in [tex]\((x^a + x^{a+1})\)[/tex].
3. First, distribute [tex]\(x^{a+2}\)[/tex]:
[tex]\[ x^{a+2} \cdot x^a + x^{a+2} \cdot x^{a+1} = x^{(a+2)+a} + x^{(a+2)+(a+1)} = x^{2a+2} + x^{2a+3} \][/tex]
4. Next, distribute [tex]\(-2x^a\)[/tex]:
[tex]\[ -2x^a \cdot x^a + -2x^a \cdot x^{a+1} = -2x^{a+a} + -2x^a \cdot x^{a+1} = -2x^{2a} + -2x^{a+(a+1)} = -2x^{2a} - 2x^{2a+1} \][/tex]
5. Finally, distribute [tex]\(3^{a+1}\)[/tex]:
[tex]\[ 3^{a+1} \cdot x^a + 3^{a+1} \cdot x^{a+1} = 3^{a+1}x^a + 3^{a+1}x^{a+1} \][/tex]
6. Combine all the expanded terms:
[tex]\[ x^{2a+2} + x^{2a+3} - 2x^{2a} - 2x^{2a+1} + 3^{a+1} x^a + 3^{a+1}x^{a+1} \][/tex]
So, the simplified expression is:
[tex]\[ 33^ax^a + 33^ax^{(a+1)} - 2x^{2a} - 2x^ax^{(a+1)} + x^ax^{(a+2)} + x^{(a+1)}x^{(a+2)} \][/tex]
Rewriting it:
[tex]\[ 33^ax^a + 33^ax^{a + 1} - 2x^{2a} - 2x^{a}* x^{a + 1} + x^{a + 1 + a + 1} + x^{a + 2 + a + 1} \][/tex]
Therefore, the expanded form of the given expression [tex]\(\left(x^{a+2} - 2x^a + 3^{a+1}\right)\left(x^a + x^{a+1}\right)\)[/tex] is given by:
[tex]\[ 3 \cdot 3^a \cdot x^a + 3 \cdot 3^a \cdot x^{a+1} - 2 \cdot x^{2a} - 2 \cdot x^{a} \cdot x^{a+1} + x^{a} \cdot x^{a+2} + x^{a+1} \cdot x^{a+2} \][/tex]