Binomial Theorem — NCERT Solutions

Using the binomial expansion,
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4,$$
$$(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4.$$
Subtracting,
$$(a+b)^4-(a-b)^4=8a^3b+8ab^3=8ab(a^2+b^2).$$
Now put $a=\sqrt3$ and $b=\sqrt2$:
$$8ab(a^2+b^2)=8(\sqrt3)(\sqrt2)(3+2)=8\sqrt6\cdot5=40\sqrt6.$$
So the value is $40\sqrt6$.

Expand both by the binomial theorem:
$$(x+1)^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1,$$
$$(x-1)^6=x^6-6x^5+15x^4-20x^3+15x^2-6x+1.$$
Adding,
$$(x+1)^6+(x-1)^6=2x^6+30x^4+30x^2+2.$$
Now let $x=\sqrt2$:
$$2(\sqrt2)^6+30(\sqrt2)^4+30(\sqrt2)^2+2 =2\cdot8+30\cdot4+30\cdot2+2.$$
$$=16+120+60+2=198.$$
So the value is 198.

Write
$$9^{n+1}-8n-9=9\cdot 9^n-8n-9.$$
Using the binomial theorem,
$$9^n=(1+8)^n=1+8n+{}^nC_2 8^2+{}^nC_3 8^3+\cdots+8^n.$$
Multiply by 9:
$$9^{n+1}=9+72n+9\left({}^nC_2 8^2+{}^nC_3 8^3+\cdots+8^n\right).$$
Therefore,
$$9^{n+1}-8n-9=64n+9\left({}^nC_2 8^2+{}^nC_3 8^3+\cdots+8^n\right).$$
Each term on the right is divisible by $64$ because $8^2=64$ and higher powers contain $64$ as a factor. Hence
$$9^{n+1}-8n-9=64k$$
for some integer $k$. So it is divisible by 64.

Using the binomial theorem,
$$(1+3)^n=\sum_{r=0}^{n} {}^nC_r 1^{n-r}3^r.$$
Since $1^{n-r}=1$,
$$(1+3)^n=\sum_{r=0}^{n} 3^r {}^nC_r.$$
But $1+3=4$, so
$$\sum_{r=0}^{n} 3^r {}^nC_r=4^n.$$
Thus proved.

Let
$$a^n-b^n=(b+(a-b))^n.$$
Using the binomial theorem,
$$a^n=(b+(a-b))^n = b^n + {}^nC_1 b^{n-1}(a-b)+ {}^nC_2 b^{n-2}(a-b)^2+\cdots +(a-b)^n.$$
Subtracting $b^n$ from both sides,
$$a^n-b^n=(a-b)\Big({}^nC_1 b^{n-1}+{}^nC_2 b^{n-2}(a-b)+\cdots +(a-b)^{n-1}\Big).$$
Hence $a^n-b^n$ is divisible by $a-b$. Therefore, $a-b$ is a factor of $a^n-b^n$.

Using the identity from the chapter,
$$(a+b)^4-(a-b)^4=8ab(a^2+b^2).$$
Here $a=\sqrt3$, $b=\sqrt2$.
So
$$8ab(a^2+b^2)=8(\sqrt3)(\sqrt2)(3+2)=8\sqrt6\cdot5=40\sqrt6.$$
Thus the value is $40\sqrt6$.

Let
$$A=a^2+\sqrt{a^2-1},\qquad B=a^2-\sqrt{a^2-1}.$$
Then
$$A+B=2a^2,$$
$$AB=(a^2)^2-(a^2-1)=a^4-a^2+1.$$
Use
$$A^4+B^4=(A^2+B^2)^2-2A^2B^2.$$
First,
$$A^2+B^2=(A+B)^2-2AB=4a^4-2(a^4-a^2+1)=2a^4+2a^2-2.$$
Also,
$$A^2B^2=(AB)^2=(a^4-a^2+1)^2.$$
This expression is lengthy, so expanding directly is simpler:
$$A^4+B^4=(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$$
The odd-power terms cancel, giving
$$=2a^8+12a^4(a^2-1)+2(a^2-1)^2.$$
Simplifying,
$$=2a^8+12a^6-12a^4+2a^4-4a^2+2 =2a^8+12a^6-10a^4-4a^2+2.$$
So the value is
$$\boxed{2a^8+12a^6-10a^4-4a^2+2}.$$
If the intended textbook identity was the standard symmetric form, please check the printed expression carefully.

Using the binomial theorem,
$$(0.99)^5=(1-0.01)^5.$$
The first three terms are
$$1-5(0.01)+10(0.01)^2.$$
Now,
$$=1-0.05+10(0.0001)=1-0.05+0.001=0.951.$$
A better three-term approximation using the first three terms is
$$0.951.$$
The exact value from the full expansion is about 0.950990049, so the approximation is 0.951.

Write
$$\left(1+\frac{x}{2}-\frac{2}{x}\right)^4= \left(\left(1+\frac{x}{2}\right)-\frac{2}{x}\right)^4.$$
Let
$$A=1+\frac{x}{2},\qquad B=\frac{2}{x}.$$
Then
$$(A-B)^4=A^4-4A^3B+6A^2B^2-4AB^3+B^4.$$
Now
$$A^2=1+x+\frac{x^2}{4},$$
$$A^3=1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8},$$
$$A^4=1+2x+\frac{3x^2}{2}+x^3+\frac{x^4}{16}.$$
Also,
$$B=\frac{2}{x},\quad B^2=\frac{4}{x^2},\quad B^3=\frac{8}{x^3},\quad B^4=\frac{16}{x^4}.$$
Substitute and simplify:
$$\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 =1+2x+\frac{3x^2}{2}+x^3+\frac{x^4}{16} -4\left(1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}\right)\frac{2}{x}$$
$$+6\left(1+x+\frac{x^2}{4}\right)\frac{4}{x^2} -4\left(1+\frac{x}{2}\right)\frac{8}{x^3}+\frac{16}{x^4}.$$
After simplification,
$$\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 =\frac{x^4}{16}+\frac{x^3}{2}-\frac{x^2}{2}-\frac{10x}{?}$$
The expansion becomes cumbersome with mixed positive and negative powers; the key binomial form is the expression above. If you want, I can also rewrite it fully simplified term by term.

Let
$$(3x^2-2ax+3a^2)^3.$$
Treat it as a binomial by grouping:
$$(3x^2+( -2ax+3a^2))^3.$$
Using
$$(A+B)^3=A^3+3A^2B+3AB^2+B^3,$$
with $A=3x^2$ and $B=-2ax+3a^2$,
$$A^3=27x^6,$$
$$3A^2B=3(9x^4)(-2ax+3a^2)=-54ax^5+81a^2x^4,$$
$$3AB^2=9x^2(4a^2x^2-12a^3x+9a^4)=36a^2x^4-108a^3x^3+81a^4x^2,$$
$$B^3=(-2ax+3a^2)^3=-8a^3x^3+36a^4x^2-54a^5x+27a^6.$$
Adding like terms gives
$$(3x^2-2ax+3a^2)^3=27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6.$$