Solve (2m-n)^2-169(m+n)^2

$$( 2 m - n ) ^ { 2 } - 169 ( m + n ) ^ { 2 }$$

Evaluate

$$3\left(-11m-14n\right)\left(5m+4n\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-n\right)^{2}.

$$4m^{2}-4mn+n^{2}-169\left(m+n\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$4m^{2}-4mn+n^{2}-169\left(m^{2}+2mn+n^{2}\right)$$

Use the distributive property to multiply -169 by m^{2}+2mn+n^{2}.

$$4m^{2}-4mn+n^{2}-169m^{2}-338mn-169n^{2}$$

Combine 4m^{2} and -169m^{2} to get -165m^{2}.

$$-165m^{2}-4mn+n^{2}-338mn-169n^{2}$$

Combine -4mn and -338mn to get -342mn.

$$-165m^{2}-342mn+n^{2}-169n^{2}$$

Combine n^{2} and -169n^{2} to get -168n^{2}.

$$-165m^{2}-342mn-168n^{2}$$

Expand

$$-165m^{2}-342mn-168n^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-n\right)^{2}.

$$4m^{2}-4mn+n^{2}-169\left(m+n\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$4m^{2}-4mn+n^{2}-169\left(m^{2}+2mn+n^{2}\right)$$

Use the distributive property to multiply -169 by m^{2}+2mn+n^{2}.

$$4m^{2}-4mn+n^{2}-169m^{2}-338mn-169n^{2}$$

Combine 4m^{2} and -169m^{2} to get -165m^{2}.

$$-165m^{2}-4mn+n^{2}-338mn-169n^{2}$$

Combine -4mn and -338mn to get -342mn.

$$-165m^{2}-342mn+n^{2}-169n^{2}$$

Combine n^{2} and -169n^{2} to get -168n^{2}.

$$-165m^{2}-342mn-168n^{2}$$